Course: 11.7 使用矢量值函数描述投影动作

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使用矢量值函数描述投影动作
Jill’s gotten a work-study job at her college golf course. One day on her lunch break she decides to see how far she can hit a golf ball. If the ball leaves the tee going 58 meters per second at a 45 degree angle, what equation can describe its flight? How far will it go before it hits the ground?
::吉尔在大学高尔夫球场获得了一份工作研究工作。午餐休息的一天,她决定看她能打高尔夫球多远。如果球在45度角度下将每秒58米的球离开球队,那么什么方程可以描述它的飞行? 在它落地之前,它能走多远?

Projectile Motion
::投射动作

You can use vector-valued functions, their , and their integrals to model the motion of projectiles. When someone fires a projectile, its initial velocity can be represented by a vector where the magnitude is the overall speed and the angle is the direction of the projectile. After its launch, a projectile’s motion can be divided into vertical and horizontal components. For simple physics problems, you assume that the horizontal component of the velocity remains constant, while the acceleration of gravity affects the vertical component .
::您可以使用矢量价值的函数、函数和元件来模拟射弹的运动。当有人发射射弹时, 其初始速度可以用矢量表示, 该矢量是总速度, 角度是射弹的方向。发射后, 射弹的运动可以分为垂直和水平部分。对于简单的物理问题, 您假设速度的水平部分保持不变, 而重力加速会影响垂直部分。

If you know the original velocity and the rate of acceleration, it’s possible to create a vector-valued function that describes the object’s position in space as a vector-valued function of time. You can use this function to find the peak height or the point when an object hits the ground. Vector-valued functions make these calculations fairly simple.
::如果您知道原始速度和加速率, 就可以创建一个矢量估值函数, 将天体在空间的位置描述为矢量估值时间函数。您可以使用此函数查找物体撞击地面时的峰值高度或点。矢量估值函数使这些计算简单化。

In the Roman Coliseum, a gladiator throws his net at an opponent across the arena. The net leaves his hand when it’s 2 meters above the ground. It starts moving at 5 m/s at a 60 degree angle. Let's find the vector-valued function that describes its position as a function of $\begin{align*}t\end{align*}$ .
::在罗马竞技场中,角斗士向竞技场另一端的对手投掷自己的网。当网在地面上两米处时,网就离开他的手。它以60度角以5米/秒的速度开始移动。让我们找到矢量价值的函数,该函数将它的位置描述为 t 的函数。

First, write the vector-valued function for the acceleration of the net. Assume that the horizontal velocity is constant. You can use $\begin{align*}-9.8 \ m/s^2\end{align*}$ as the acceleration due to gravity. That means that $\begin{align*}\vec{A}(t) = (0, -9.8)\end{align*}$ .
::首先,为网络加速度写入矢量估值函数。假设水平速度不变。您可以使用-9.8 m/s2作为重力加速度。这意味着 A(t)=(0)-9.8。

Velocity is the integral of acceleration. So,
::速度是加速的有机组成部分。所以,

$\begin{aligned} \vec{V} (t) = \int \vec{A} (t) d t \\ \vec{V} (t) = (0 t + C, - 9.8 t + K) \end{aligned}$
$\begin{align*}\vec{V}(t) = \int \vec{A}(t) dt \\ \vec{V}(t) = (0t + C, -9.8t + K)\end{align*}$
::V(t)A(t)dtv(t)=(0t+C,-9.8t+K)

When you’re dealing with a vector-valued function you can think of the constants as a vector . In this case, represents the original velocity of the net. You can compute $\begin{align*}(C, K)\end{align*}$ from the magnitude and angle of the initial velocity by finding the $\begin{align*}i\end{align*}$ and $\begin{align*}j\end{align*}$ components of the vector.
::当您正在处理矢量估值函数时, 您可以将常数视为矢量。在此情况下, 表示网络的原始速度。您可以从初始速度的大小和角度通过找到矢量的 i 和 j 组件来计算( C, K ) 。

$\begin{aligned} ‖ v_{0} ‖ = 5, θ = 60^{\circ} \\ \cos θ = \frac{i}{‖ v_{0} ‖}, \sin θ = \frac{j}{‖ v_{0} ‖} \\ \cos (60) = \frac{i}{5} \\ i = \frac{5}{2} \\ \sin (60) = \frac{j}{5} \\ j = \frac{5 \sqrt{3}}{2} \end{aligned}$
$\begin{align*}\Big \| v_0 \Big \| = 5, \theta = 60^\circ \\ \cos \theta = \frac{i}{\Big \| v_0 \Big \|}, \sin \theta = \frac{j}{\Big \| v_0\Big \| } \\ \cos(60) = \frac{i}{5} \\ i=\frac{5}{2} \\ \sin(60) = \frac{j}{5} \\ j = \frac{5 \sqrt{3}}{2}\end{align*}$
::566666555555555555555

So, the initial velocity can be described by the vector $\begin{align*}\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)\end{align*}$ and the vector-valued function that describes the velocity of the net as a function of time is $\begin{align*}\vec{V}(t) = \left(\frac{5}{2}, -9.8t + \frac{5 \sqrt{3}}{2}\right)\end{align*}$ .
::因此,初始速度可以用矢量(52,532)和矢量估值函数来描述初始速度,该函数用时间函数来描述网络的速度,即V(t)=(52,-9.8t+532)。

Velocity is the rate at which distance changes. This means that you can find the vector-valued function for the position of the net by taking the integral of the vector-valued function for velocity.
::速度是距离变化的速率。这意味着您可以通过以矢量估值函数的集成值计算速度来找到网络位置的矢量估值函数。

$\begin{aligned} \vec{D} (t) = \int \vec{V} (t) d t \\ \vec{D} (t) = (\frac{5}{2} t + C, \frac{- 9.8}{2} t^{2} + \frac{5 \sqrt{3}}{2} t + K) \end{aligned}$
$\begin{align*}\vec{D}(t) = \int \vec{V} (t) dt \\ \vec{D}(t) = \left(\frac{5}{2}t+C, \frac{-9.8}{2}t^2 + \frac{5 \sqrt{3}}{2}t+K \right)\end{align*}$
::D(t)V(t)dtD(t)=(52t+C,-9.82t2+532t+K)

In this case, the constant vector (C, K) will be the starting position of the net. The gladiator released the net at a height of 2 meters, so the starting position for the net is $\begin{align*}(0, 2)\end{align*}$ . Your vector-valued function for the distance that the net has traveled with respect to time will be:
::在此情况下, 恒定矢量 (C, K) 将是网的起始位置。角斗士在 2 米高的高度释放网, 所以网的起始位置是 (0, 2) 。您的矢量值函数是网在时间上行走的距离 :

$\begin{aligned} \vec{D} (t) = (\frac{5}{2} t, \frac{- 9.8}{2} t^{2} + \frac{5 \sqrt{3}}{2} t + 2) . \end{aligned}$
$\begin{align*}\vec{D} (t) = \left(\frac{5}{2}t, \frac{-9.8}{2}t^2 + \frac{5 \sqrt{3}}{2} t+2 \right ).\end{align*}$
::D(t) =(52t,-9.82t2+532t+2)。

If you’ve studied physics, you may notice how much the vertical path of this vector-valued function resembles the equation for distance: $\begin{align*}d = v_0 t + \frac{1}{2}at^2\end{align*}$ . The difference is that the equation you memorized when you first began to study physics only dealt with movement in one dimension. Vector-valued functions allow you to apply the principles of mechanics in two or more dimensions.
::如果您研究过物理, 您可能会注意到这个矢量值函数的垂直路径与距离公式( d=v0t+12at2) 的相似程度。区别在于, 当您刚开始研究物理时, 所记住的方程只涉及一个维度的移动。矢量值函数允许您在两个或多个维度中应用机械原理。

Once you have vector-valued functions that describe a projectile’s acceleration, velocity and position, you can use them to find more information about that object’s flight. For instance, you can find the maximum height that the gladiator’s net reached, and at what time it reached that height.
::一旦您有了描述投射物加速度、速度和位置的矢量估值函数,您就可以使用这些函数查找更多关于该物体飞行的信息。比如,您可以找到角斗士网达到的最高高度,以及它达到该高度的时间。

Because the motion of a projectile is affected by gravity, its vertical velocity will decrease as the acceleration of gravity affects its flight. The vertical velocity will be zero at the peak of its arc, just before the object begins to return to Earth with a negative velocity.
::由于射弹的运动受到重力的影响,其垂直速度将随着重力加速影响其飞行而下降。在物体开始以负速返回地球之前,垂直速度将在其弧的峰值时为零。

So, to find the peak height for the gladiator’s net, you’ll need to find the time at which the vertical velocity is 0.
::因此,要找到角斗士网的峰值高度, 您需要找到垂直速度为 0 的时间。

$\begin{aligned} \vec{V} (t) = (\frac{5}{2}, - 9.8 t + \frac{5 \sqrt{3}}{2}) \\ - 9.8 t + \frac{5 \sqrt{3}}{2} = 0 \\ - 9.8 t = - \frac{5 \sqrt{3}}{2} \\ t = .44 \end{aligned}$
$\begin{align*}\vec{V}(t) = \left(\frac{5}{2}, -9.8t + \frac{5 \sqrt{3}}{2} \right) \\ -9.8t + \frac{5 \sqrt{3}}{2} = 0 \\ -9.8t = -\frac{5 \sqrt{3}}{2} \\ t = .44\end{align*}$
::V(t) =(52,-9.8t+532)-9.8t+532-9.8t+532=0-9.8t=0.532=0-9.8t532t=44

At .44 seconds into its flight, the net will reach its peak height. To find out what height the net reaches, evaluate the vector-valued function that expresses distance when $\begin{align*}t=.44\end{align*}$ .
::在飞行44秒后,网将达到峰值高度。要了解网的高度,请评估在 t=44时显示距离的矢量值函数。

$\begin{aligned} \vec{D} (.44) = (\frac{5}{2} (.44), \frac{- 9.8}{2} (.44)^{2} + \frac{5 \sqrt{3}}{2} (.44) + 2) \\ \vec{D} (.44) = (1.1, 2.96) \end{aligned}$
$\begin{align*}\vec{D} (.44) = \left(\frac{5}{2} (.44), \frac{-9.8}{2} (.44)^2 + \frac{5 \sqrt{3}}{2} (.44) + 2 \right) \\ \vec{D} (.44) = (1.1,2.96)\end{align*}$
::D_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

The net will reach its peak at a height of 2.96 meters. It will reach this peak 1.1 meters away from the gladiator.
::网顶将达到2.96米的高度,达到距离角斗士1.1米的峰值。

You can also use vector-valued functions to determine when and where a projectile will strike the ground. The projectile strikes the ground when the vertical component of the distance function is equal to zero. When and where does the gladiator's net hit the ground?
::您也可以使用矢量价值的函数来确定弹丸何时和在哪里撞击地面。当距离函数的垂直部分等于零时,弹丸会撞击地面。角斗士的净额何时和在哪里击中地面?

If you go back to the gladiator’s net, you can see that the vertical component of the distance function is
::如果您回到角斗斗士的网网,可以看到距离函数的垂直部分是

$\begin{aligned} 0 = \frac{- 9.8}{2} t^{2} + \frac{5 \sqrt{3}}{2} t + 2 \end{aligned}$
$\begin{align*}0=\frac{-9.8}{2}t^2 + \frac{5 \sqrt{3}}{2} t+2\end{align*}$
::09.82t2+532t+2

Use the quadratic formula to solve, and you’ll find that
::使用四方公式解决,你会发现

$\begin{aligned} 0 = \frac{- 9.8}{2} t^{2} + \frac{5 \sqrt{3}}{2} t + 2 \\ \frac{- \frac{5 \sqrt{3}}{2} \pm \sqrt{{(\frac{5 \sqrt{3}}{2})}^{2} - 4 (\frac{- 9.8}{2})} 2}{2 (\frac{- 9.8}{2})} = 1.2186 \end{aligned}$
$\begin{align*}0 = \frac{-9.8}{2}t^2 + \frac{5 \sqrt{3}}{2} t + 2 \\ \frac{-\frac{5 \sqrt{3}}{2} \pm \sqrt{\left(\frac{5 \sqrt{3}}{2}\right)^2 -4 \left(\frac{-9.8}{2}\right)}2} {2 \left(\frac{-9.8}{2}\right)} = 1.2186\end{align*}$
::09.82t2+532t+2-532(532)2-4(9.82)22(-9.82)=1.2186

Note that in this case it doesn't make sense to consider the negative answer of -0.3349 because time must be positive. So, the net lands when 1.2186 seconds have passed. To find out how far the net flies, evaluate the horizontal portion of the distance function when $\begin{align*}t=1.2186\end{align*}$ .
::请注意, 在此情况下, 考虑 - 0. 3349 的否定回答是没有意义的, 因为时间必须是正的。因此, 当 1.2186 秒已经过时, 净陆地已经是正的。要了解净苍蝇的距离, 请在 t= 1. 2186 时评估距离函数的横向部分。

$\begin{aligned} \vec{D} (t) = (\frac{5}{2} t, \frac{- 9.8}{2} t^{2} + \frac{5 \sqrt{3}}{2} t + 2) \\ \frac{5}{2} (.82) = 3.05 \end{aligned}$
$\begin{align*}\vec{D} (t) = \left(\frac{5}{2}t, \frac{-9.8}{2} t^2 + \frac{5 \sqrt{3}}{2} t + 2 \right) \\ \frac{5}{2}(.82) = 3.05\end{align*}$
::D(t) =(52t,-9.82t2+532t+2) 52(82) =3.05

The net will hit the ground 3.05 meters away from the gladiator.
::网将击中距离角斗士3.05米的地面

Examples
::实例

Example 1
::例1

Earlier, you were asked to find an equation to describe the movement of Jill's golf ball and to calculate when it will hit the ground. Jill knows that the force of gravity will act on her golf ball at $\begin{align*}-9.8 \ m/s^2\end{align*}$ , so the vector-valued function representing acceleration is (0, -9.8).
::早些时候,有人要求你找到一个方程来描述吉尔的高尔夫球的动向,并计算它何时撞击地面。吉尔知道重力将在-9.8 m/s2对她的高尔夫球起作用,所以代表加速度的矢量值函数是(0,9.8)。

To find the vector-valued function for velocity, she’ll need to write the ball’s starting velocity in terms of $\begin{align*}i\end{align*}$ and $\begin{align*}j\end{align*}$ . She can use the definitions of sine and cosine to break the vector into its component parts.
::为了找到速度的矢量值函数, 她需要用 i 和j 来写出球的起始速度。她可以使用正弦和余弦的定义来将矢量断裂成其组件部分。

The magnitude of her initial velocity vector was 58 m/s and the angle of the launch was 45 degrees, so
::她最初速度矢量的高度为58米/秒,发射角度为45度,所以

$\begin{aligned} ‖ v_{0} ‖ = 58, θ = 45^{\circ} \\ \cos 45 = \frac{i}{58} \\ i = 29 \sqrt{2} \\ \sin 45 = \frac{j}{58} \\ j = 29 \sqrt{2} \end{aligned}$
$\begin{align*}\Big \| v_0 \Big \| = 58, \theta=45^\circ \\ \cos 45 = \frac{i}{58} \\ i = 29 \sqrt{2} \\ \sin 45 = \frac{j}{58} \\ j = 29 \sqrt{2}\end{align*}$
::58 45 45 45=i58i=292sin=45j58j=292

The initial velocity of the golf ball can be described by the vector $\begin{align*}\left(29 \sqrt{2}, 29 \sqrt{2} \right)\end{align*}$ .
::矢量可以描述高尔夫球的最初速度(292,292)。

The vector-valued function that describes the velocity of the golf ball is:
::说明高尔夫球速度的矢量估值函数是:

$\begin{aligned} \vec{V} (t) = (29 \sqrt{2}, 29 \sqrt{2} - 9.8 t) \end{aligned}$
$\begin{align*}\vec{V}(t) = \left(29 \sqrt{2}, 29 \sqrt{2} - 9.8t \right)\end{align*}$
::V(t) = (292,292-9.8t)

To find the vector-valued function $\begin{align*}\vec{D} (t)\end{align*}$ that describes the ball’s position, Jill can take the integral of the velocity function. If she assumes the initial position of her golf ball is (0, 0), she’ll find that $\begin{align*}\vec{D}(t) = \left(29 \sqrt{2t}, 29 \sqrt{2t} - \frac{9.8}{2}t^2 \right)\end{align*}$ . She can use these vector-valued functions to find out how far her ball will go.
::为了找到描述球位置的矢量价值函数 D( t) , Jill 可以选择速度函数的有机体。如果她假设她的高尔夫球最初位置是 0, 0 , 她会发现 D( t) = ( 292t, 292t-9.82t.2) 。她可以使用这些矢量价值函数来了解她的球能走多远。

The golf ball will hit the ground when the vertical component of its position vector is equal to zero.
::当高尔夫球位置矢量的垂直部分等于零时,高尔夫球将击中地面。

$\begin{aligned} 0 = 29 \sqrt{2} t - \frac{9.8}{2} t^{2} \end{aligned}$
$\begin{align*}0=29\sqrt{2} t - \frac{9.8}{2} t^2\end{align*}$
::0=292t-9.82t2

Jill can use the quadratic formula to find the time at which the ball hits the ground.
::吉尔可以使用二次方程式找到球击中地面的时间

$\begin{aligned} t = \frac{- 29 \sqrt{2} \pm \sqrt{{(29 \sqrt{2})}^{2} - 4 (- \frac{9.8}{2}) (0)}}{2 (- \frac{9.8}{2})} \\ t = 0 \\ t = 8.37 \end{aligned}$
$\begin{align*}t = \frac{-29 \sqrt{2} \pm \sqrt{\left(29 \sqrt{2}\right)^2 -4 \left(-\frac{9.8}{2}\right)(0)}}{2 \left(-\frac{9.8}{2}\right)} \\ t=0 \\ t=8.37\end{align*}$
::t292(292)2-4(-9.82)(0)(2-9.82t=0t=8.37)

The ball hits the ground after 8.37 seconds. Substitute this time back into the distance equation to find the horizontal distance that her ball travels.
::球在8.37秒后撞击到地面。替换这个时间返回到距离方程中, 以找到她的球所穿梭的水平距离。

$\begin{aligned} \vec{D}^{'} (t) = (29 \sqrt{2} t, 29 \sqrt{2} t - \frac{9.8}{2} t^{2}) \\ 29 \sqrt{2} (8.37) = 343.27 \end{aligned}$
$\begin{align*}\vec{D}{^\prime}(t) = \left ( 29 \sqrt{2}t, 29 \sqrt{2}t - \frac{9.8}{2}t^2 \right )\\ 29 \sqrt{2} (8.37) = 343.27\end{align*}$
::D(t)=(292,292t-9.82t2)292(8.37)=343.27。

Her ball travels 343.27 meters from the tee.
::她的球距离球队343.27米

An astronaut travels to the purple planet, where the acceleration due to gravity is $\begin{align*}4 \ m/s^2\end{align*}$ . The strange, murky atmosphere on the planet causes projectiles to decelerate by $\begin{align*}.5 \ m/s\end{align*}$ every second they’re in flight. He kicks a soccer ball at $\begin{align*}14 \ m/s\end{align*}$ at an angle of 30 degrees. For the following examples, use this information.
::宇航员前往紫色星球,重力加速度为4 m/s2. 地球的奇特、黑暗的大气层每飞行一秒就导致射弹减速0.5 m/s。他将球踢在14 m/s,角度为30 度。以下例子请使用这些信息。

Example 2
::例2

Write vector-valued functions to describe the velocity and position of the soccer ball as it flies.
::写入矢量估值功能,以描述足球飞行的速度和位置。

First, find the function for acceleration. To find the function for velocity you can integrate the function for velocity.
::首先,找到加速的函数。要找到速度的函数,您可以整合速度的函数。

$\begin{aligned} \vec{A} (t) = (- .5, - 4) \\ \vec{V} (t) = \int \vec{A} (t) d t + v_{0} \end{aligned}$
$\begin{align*}\vec{A}(t) = (-.5, -4) \\ \vec{V}(t) = \int \vec{A}(t) dt + v_0\end{align*}$
::A(t) = (-5,-4) V(t) A(t) dt+v0

Find the $\begin{align*}i\end{align*}$ and $\begin{align*}j\end{align*}$ components of the original velocity vector.
::查找原始速度矢量的 I 和 j 组件。

$\begin{aligned} \cos (30) = \frac{i}{14} \\ i = 7 \sqrt{3} \\ \sin (30) = \frac{j}{14} \\ j = 7 \\ v_{0} = (7 \sqrt{3}, 7) \end{aligned}$
$\begin{align*}\cos(30) = \frac{i}{14} \\ i = 7 \sqrt{3} \\ \sin(30) = \frac{j}{14} \\ j=7 \\ v_0 = \left(7 \sqrt{3}, 7 \right)\end{align*}$
:30)=i14i=73sin(30)=j14j=7v0=(73,7)

Now you can find the velocity function.
::现在你可以找到速度函数了

$\begin{aligned} \vec{V} (t) = \int \vec{A} (t) d t + v_{0} \\ \vec{V} (t) = (\int - .5 d t \int - 4 d t) + (7 \sqrt{3}, 7) \\ \vec{V} (t) = (- .5 t, - 4 t) + (7 \sqrt{3}, 7) \\ \vec{V} (t) = (- .5 t + 7 \sqrt{3}, - 4 t + 7) \end{aligned}$
$\begin{align*}\vec{V}(t) = \int \vec{A}(t) dt + v_0 \\ \vec{V}(t) = \left(\int -.5 dt \int -4dt \right) + \left(7 \sqrt{3}, 7 \right) \\ \vec{V}(t) = (-.5t, -4t) + \left(7 \sqrt{3}, 7 \right) \\ \vec{V}(t) = \left(-.5t + 7 \sqrt{3}, -4t + 7 \right) \\\end{align*}$
::V(t) A(t) d(t) +v00V(t) = (5d4dt) +(73,7)V(t) =(-5,7) +(73,7) V(t) =(-5t+73,-4t+7)

Find the integral of $\begin{align*}\vec{V}(t)\end{align*}$ in order to find the function $\begin{align*}\vec{D}(t)\end{align*}$ . Based on the original problem, you can assume that the soccer ball’s starting position was (0, 0).
::查找 V( t) 的积分以查找函数 D( t) 。基于最初的问题, 您可以假设足球的起始位置是 0, 0 。

$\begin{aligned} \vec{D} (t) = \int \vec{V} (t) d t \\ \vec{D} (t) = (\int - .5 t + 7 \sqrt{3} d t, \int - 4 t + 7 d t) \\ \vec{D} (t) = (- .25 t^{2} + 7 \sqrt{3} t, - 2 t^{2} + 7 t) \end{aligned}$
$\begin{align*}\vec{D}(t) = \int \vec{V}(t) dt \\ \vec{D}(t) = \left(\int -.5t + 7 \sqrt{3} dt, \int -4t + 7 dt \right) \\ \vec{D}(t) = \left(-.25t^2 + 7 \sqrt{3}t, -2t^2 + 7t \right) \\\end{align*}$
::D(t) V(t) d(t) = (.5t+73dt,4t+7dt) D(t) = (- 25t2+73t,-2t2+7t)

Example 3
::例3

When does the soccer ball reach the top of its arc? At what position does it peak?
::足球球何时到达其弧顶端? 它在什么位置达到峰值?

The soccer ball will reach the top of its arc when the vertical velocity is zero.
::当垂直速度为零时,足球球将到达其弧的顶部。

$\begin{aligned} 0 = - 4 t + 7 \\ \frac{7}{4} = t \end{aligned}$
$\begin{align*}0 = -4t + 7 \\ \frac{7}{4} = t\end{align*}$
::04t+774=t

Plug this time into the height component of the distance equation in order to find the maximum height.
::将这个时间插到距离方程的高度组成部分中, 以便找到最大高度。

$\begin{aligned} (- 2 {(\frac{7}{4})}^{2}) + \frac{7}{4} (7) = h e i g h t \\ - \frac{49}{8} + \frac{49}{4} = h e i g h t \\ \frac{49}{8} = 6.13 m \end{aligned}$
$\begin{align*}\left(-2 \left(\frac{7}{4}\right)^2 \right) + \frac{7}{4}\left(7\right) = height \\ -\frac{49}{8} + \frac{49}{4} = height \\ \frac{49}{8} = 6.13 \ m \end{align*}$
:-2,742)+74(7)=高度-498+494=高度498=6.13米)

The soccer ball reaches its peak 6.13 m off the ground. To find the horizontal distance it travels to reach its peak, evaluate the horizontal portion of the function at $\begin{align*}t=\frac{7}{4}\end{align*}$ .
::足球在离地面6.13米处达到峰值。要找到它到达峰值的水平距离,请在 t=74 时评估函数的水平部分。

$\begin{aligned} - .25 {(\frac{7}{4})}^{2} + 7 \sqrt{3} (\frac{7}{4}) = 20.45 \end{aligned}$
$\begin{align*}-.25 \left(\frac{7}{4}\right)^2 + 7 \sqrt{3} \left(\frac{7}{4}\right) = 20.45\end{align*}$

The ball will have traveled 20.45 meters when it reaches the peak of its arc.
::当球到达其弧的峰值时,它会行走20.45米。

Example 4
::例4

Where does the soccer ball land? At what time does it land?
::足球在哪里落地?

The soccer ball will land when its vertical position is 0. So to find the time at which it lands, find out when the vertical component of the vector-valued function for distance is 0.
::当足球的垂直位置为 0 时, 足球会降落。所以要找到它降落的时间, 找出矢量值的距离函数的垂直部分是 0 。

$\begin{aligned} - 2 t^{2} + 7 t = 0 \\ t = \frac{- 7 \pm \sqrt{7^{2} - 4 (2) (0)}}{2 (- 2)} \\ t = \frac{7}{2} o r = 0 \end{aligned}$
$\begin{align*}-2t^2 + 7t = 0 \\ t = \frac{-7 \pm \sqrt{7^2 - 4(2)(0)}}{2(-2)} \\ t = \frac{7}{2} \ or \ = 0\end{align*}$
::− 2t2+7t=0t772-4(2)(0)(2-2-2)-2t=72或=0

The ball is on the ground when $\begin{align*}t=0\end{align*}$ (which you knew from the problem) and when $\begin{align*}t=\frac{7}{2}\end{align*}$ . Find the horizontal position at $\begin{align*}t=\frac{7}{2}\end{align*}$ by evaluating the horizontal position of the distance equation for $\begin{align*}t=\frac{7}{2}\end{align*}$ .
::当 t=0 (您从问题中知道) 和 t=72 时,球在地面上。通过评价 t=72 的距离方程水平位置,找到 t=72 的水平位置。

$\begin{aligned} - .25 {(\frac{7}{2})}^{2} + 7 \sqrt{3} (\frac{7}{2}) = 39.37 \end{aligned}$
$\begin{align*}-.25 \left ( \frac{7}{2} \right )^2 + 7\sqrt{3} \left ( \frac{7}{2} \right ) = 39.37\end{align*}$

The ball will travel 39.37 meters before it hits the ground on the purple planet.
::球在进入紫色星球之前将行走39.37米

Review
::回顾

For #1-5, use the following information:
::对于第1-5号,请使用以下信息:

Melody kicks a football at $\begin{align*}20 \ m/s\end{align*}$ at an angle of 42 degrees. Melody is on Earth where the acceleration due to gravity is $\begin{align*}9.8 \ m/s^2\end{align*}$ .
::Melody以20米/秒的速度踢足球,角度为42度。Melody在地球上,重力加速度为9.8米/秒。

Find the vector-valued function that models the acceleration of the football as a function of time.
::找到一个矢量值函数,该函数以时间函数来模拟足球加速率。

Find the vector-valued function that models the velocity of the football as a function of time.
::找到矢量值函数, 以时间函数来模拟足球的速度。

Find the vector-valued function that models the position of the football as a function of time.
::查找矢量值函数, 以时间函数来模拟足球的位置。

When does the football reach the top of its arc? At what position does it peak?
::足球何时到达其弧顶端?

When does the football land? How far will it have traveled?
::足球何时落地?它会走多远?

For #6-10, use the following information:
::对于第6-10号,请使用以下信息:

William tosses an eraser at $\begin{align*}7 \ m/s\end{align*}$ to his friend across the room. The eraser leaves his hand when it's 1.5 meters above the ground at a 60 degree angle to the horizontal. William is on Earth where the acceleration due to gravity is $\begin{align*}9.8 \ m/s^2\end{align*}$ .
::威廉在7m/s 向房间对面的他朋友扔一个抹布机。抹布机在距地面1.5米处, 距地面60度角, 向水平倾斜时, 将手伸出来。威廉在地球上, 重力加速度为9. 8 m/ s2 。

Find the vector-valued function that models the acceleration of the eraser as a function of time.
::查找矢量估值函数,该函数将橡皮加速度作为时间函数来模拟。

Find the vector-valued function that models the velocity of the eraser as a function of time.
::查找矢量估值函数,该函数以时间函数来模拟擦除器的速度。

Find the vector-valued function that models the position of the eraser as a function of time.
::查找矢量值函数,该函数以时间函数来模拟擦除器的位置。

When does the eraser reach the top of its arc? At what position does it peak?
::橡皮何时到达其弧顶端? 它在什么位置达到峰值 ?

Assuming William's friend does not catch the eraser and it lands on the ground, when does the eraser land? How far will it have traveled?
::假设威廉的朋友没有抓住抹布机,它会降落在地上, 抹布机何时降落?它会走多远?

For #11-15, use the following information:
::对于第11-15号,请使用以下信息:

William has moved to Mars where the acceleration due to gravity is only $\begin{align*}3.75 \ m/s^2\end{align*}$ . Once again he tosses an eraser at $\begin{align*}7 \ m/s\end{align*}$ to his friend. The eraser leaves his hand when it's 1.5 meters above the ground at a 60 degree angle to the horizontal.
::William已移动到火星, 重力加速度只有3. 75 m/ss2. 。再次, 他将一个7m/s的橡皮笔扔给他的朋友。橡皮笔在距水平60度角的地面上1.5米处离开手。

Find the vector-valued function that models the acceleration of the eraser as a function of time.
::查找矢量估值函数,该函数将橡皮加速度作为时间函数来模拟。

Find the vector-valued function that models the velocity of the eraser as a function of time.
::查找矢量估值函数,该函数以时间函数来模拟擦除器的速度。

Find the vector-valued function that models the position of the eraser as a function of time.
::查找矢量值函数,该函数以时间函数来模拟擦除器的位置。

When does the eraser reach the top of its arc? At what position does it peak?
::橡皮何时到达其弧顶端? 它在什么位置达到峰值 ?

Assuming William's friend does not catch the eraser and it lands on the ground, when does the eraser land? How far will it have traveled?
::假设威廉的朋友没有抓住抹布机,它会降落在地上, 抹布机何时降落?它会走多远?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

使用矢量值函数描述投影动作

Projectile Motion ::投射动作

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Projectile Motion
::投射动作

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)