Question

A ball is kicked at an angle . It is intended that the ball lands in 휃=44.5°the back of a moving truck which has a trunk of length L = 2 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick, is d = 7.3 m, and the truck moves directly away from the starting position with a velocity v = 11.4 m/s, find the maximum and minimum initial velocity of the ball, v0, so that the ball lands in the truck bed. Assume that the initial height of the ball is equal to the height of the ball at the instant it begins to enter the truck bed.

Answer 1

Answer:

$V_{0min}=19.63m/s$

$V_{0max}=20.45m/s$

Explanation:

In order to solve this problem, we must start by drawing a diagram of the situation, this will help us understand the problem better so we can take the best possible approach. (See attached picture).

So next, we can see that the ball will have a parabolic trajectory, so we can split it into its x- and y-components. Let's start by analyzing the movement of the ball in the x-direction.

$x_{b}=x_{0b}+V_{0bx}t$

we know that the x-component of the velocity of the ball can be found by using the following equation:

$V_{0bx}=V_{0b}cos \theta$

So the equation can be rewritten as:

$x_{b}=x_{0b}+V_{0b}(cos \theta) t$

we can say that the ball will start its movement from the origin, so the initial displacement is zero. So the simplified equation is:

$x_{b}=V_{0b}(cos \theta) t$

So now we can analyze the vertical movement of the ball. We can describe it with the equation:

$y_{b}=y_{0b}+V_{0by}t+\frac{1}{2}at^{2}$

we can say that the ball will start its movement at the origin so we can say that the initial position of ball is equal to zero and the final position is zero as well, so the equation turns to:

$0=V_{0by}t+\frac{1}{2}at^{2}$

we can describe the vertical velocity of the ball with the following equation:

$V_{0by}=V_{0b}sin \theta$

and the acceleration as:

$a=-g=-9.81m/s^{2}$

so we can substitute them in the original equation:

$(V_{0b}sin \theta)t-\frac{1}{2}gt^{2}=0$

which can be solved for t by factoring, so we get:

$t(V_{0b}sin\theta - \frac{gt}{2})=0$

so we get:

t=0 which is the initial moment at which the ball starts moving

$V_{0b}sin\theta - \frac{gt}{2}=0$

which solves to:

$t=\frac{2V_{b0}}{g}sin \theta$

so we can substitute this into the x-movement equation:

$x_{b}=V_{0b}cos \theta(\frac{2V_{b0}}{g}sin \theta)$

which simplifies to:

$x_{b}=\frac{2V_{b0}^{2}cos\theta sin\theta}{g}$

By using trigonometric identities we can further simplify this to:

$x_{b}=\frac{V_{b0}^{2}sin(2\theta)}{g}$

We can leave that equation there by the time and start analyzing the movement of the truck. We can determine the distance between the back of the truck and the initial position of the ball with the following equation:

$x_{b}=x_{t}+d$

we know that:

$x_{t}=V_{t}t$

from the previos steps we know that the time t can be found by analyzing the y-movement of the ball, so we can use that same equation to substitute for t, so we get:

$x_{t}=V_{t}(\frac{2V_{b0}}{g}sin \theta)$

which can be substituted into the first equation:

$x_{b}=V_{t}(\frac{2V_{b0}}{g}sin \theta)+d$

and we can combine it with the equation we got for the x-movement of the ball.

$V_{t}(\frac{2V_{b0}}{g}sin \theta)+d=\frac{V_{b0}^{2}sin(2\theta)}{g}$

So we are dealing with a quadratic equation which can be rewritten like this:

$\frac{2V_{t}V_{0b}sin\theta}{g}+d=\frac{V_{0b}^{2}sin(2\theta)}{g}$

we can rewrite it in standard form so we get:

$\frac{V_{0b}^{2}sin(2\theta)}{g}-\frac{2V_{t}V_{0b}sin\theta}{g}-d=0$

So we can proceed and substitute the values we know, so we get:

$\frac{V_{0b}^{2}sin(2(44.5^{o})}{9.81}-\frac{2(11.4)V_{0b}sin(44.5^{o})}{9.81}-7.3=0$

which yields:

$0.10192V_{0b}^{2}-1.629V_{0b}-7.3=0$

which can be solved by using the quadratic formula:

$V_{b0}=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

so we get:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-7.3)}}{2(0.10192)}$

which gives us two answers:

$V_{b0min}=19.63m/s$

or

$V_{b0min}=-3.65m/s$

Due to the nature of the problem we only take the positive answer, since

a negative answer would mean the ball is moving to the opposite side. So the answer is:

$V_{b0min}=19.63m/s$

In order to find the maximum velocity we follow the sme procedure, with the difference that in this case d=7.3m+2m=9.3m

The additional 2 meters come from the length of the trunk of the truck.

Turning the equation into:

$0.10192V_{0b}^{2}-1.629V_{0b}-9.3=0$

and changing the answers to:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-9.3)}}{2(0.10192)}$

$V_{b0max}=20.45m/s$

or

$V_{b0min}=-4.46m/s$

so we take only the positive answer into account leaving us with the answer:

$V_{b0max}=20.45m/s$