A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

c. For what value of h does the collision occur at the instant when the first ball is at its highest point?

Question

A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

c. For what value of h does the collision occur at the instant when the first ball is at its highest point?

Accepted Answer

Hey, everyone in this problem, a compressed spring fires vertically. A small ball from the floor with an initial speed of 6m/s simultaneously. A student located exactly at a distance D above the spring releases a box without an initial speed. We're asked to determine the position of the student above the floor at which the ball is at its maximum elevation. When it strikes the box, we're given four answer choices. A 3.67 m B 4.86 m C 9. m and D 96.2 m. Let's go ahead and draw a diagram of what we have going on. So on the ground, we have a compressed spring with a ball and the ball is going to be fired vertically with an initial speed and we're gonna call it V not B to indicate it's for the ball of six m/s. Then we have a student located at a distance D above the spring that's going to release a box. So the distance between the box, the ground here is D And they are gonna release that box without an initial speed. So v knot is equal to 0m/s. We're gonna take up to be a positive direction. And let's think about what information we know about the ball and about the box for the ball, we know the initial speed V not B is equal to six m per second, the final speed or velocity of the ball. Well, we're interested in finding the position of the student above the floor. We want the ball to be at its maximum elevation when it strikes the box. If the ball is at its maximum elevation, then its speed will be zero m per second. Ok. That ball is gonna come to rest momentarily as it goes from upwards to back downwards. So at its maximum elevation for a brief moment, it has no speed. So the final speed of the ball is zero m per second. The acceleration A B is gonna be the acceleration due to gravity. We've chosen up to be our positive direction. So the acceleration will be negative 9.8 m per second squared. We don't know the change in position, ok. The maximum elevation of the ball we don't know and we don't know the amount of time that this takes For the box. The initial speed we're given is 0m/s. We don't know the speed. The final speed when it strikes the ball, the acceleration will be the same as the ball. The acceleration due to gravity negative 9.8 m/s squared. Don't know the distance it travels. We don't know the time. Now you'll notice I've written T just as T for both the ball in the box and that's for a reason because that time is gonna be the same. We're talking about the time when the ball in the box collide, it's the same amount of time since they're released simultaneously. When we look at the displacement of the box, the total distance from the box to the ground is D, the displacement of the ball is delta Y B. So the ball is gonna have, you can imagine some maximum elevation and that is given by delta Y B. And so the distance the box travels is gonna be D minus delta Y B, which means the displacement is going to be negative and in brackets D minus delta Y B and we have that negative because we've chosen up to be the positive Y direction. The box is moving downwards. And so that is a negative displacement. OK. So in order to find the position above the floor D, we need to find delta Y, this, this information about the box, we don't have enough information about the box. We only have two known values. So let's start with the ball. If we figure out delta Y B, we can use that in our equation, we can also use the time T since we know the times are the same. So let's start by finding the maximum elevation of the ball, which is given by delta Y B. We're gonna choose the kinematic equation that includes the known variables. V not B V F B A B. And the one we're trying to find delta Y B. So we have V F B squared is equal to V, not B squared plus two multiplied by A B multiplied by delta Y B. Substituting in the values we have, we have zero on the left hand side, on the right hand side, we have six m/s squared plus two multiplied by negative 9. m per second squared multiplied by delta Y B. And if we, so for delta Y B, we have delta Y B is equal to 36 m squared per second squared divided by two times 9.8 m per second squared, Which gives us a value of 1.8, 3, 67 m. So this is our delta Y B 1.8367 m. OK? So now we have some more information about the box. OK. We know the initial speed, we know the acceleration, we don't know the displacement, but we do know part of the displacement. OK? We don't know the time. So let's use the ball information to find the time. T then we can get back to the box with three known values and we don't have three known values yet for the box. So by the time it takes to reach the maximum elevation. Now, we can choose any of the equations that include time. Here, we know all of the other variable. So let's choose the simplest one V F is equal to V nought plus a multiplied by T Subsequuting in the values we know zero is equal to six m per second plus negative 9.8 m per second squared multiplied by the time T we can solve for the time tea And we can move it to the other side and divide by negative 9.8. And we get that the time T is equal to 6m/s, divided by 9.8 m per second squared For a time of 0. seconds. And you can check with your professor on how many significant digits they like you to follow through with in your intermediate calculations. OK. So at time 0. seconds, the ball has traveled a distance of D minus delta Y B. So we can go back to the information about the box. Now, now for the box, we have the initial speed, the acceleration, the time and we wanna find the displacement because the displacement has that value of D in it. So we're gonna choose the equation without V F. We have delta Y is equal to V multiplied by T plus one half, multiplied by a multiplied by T squared, the displacement delta Y is given by negative brackets D minus delta Y B Which is 1.8, 36, 7 m and the displacement of that ball. So the maximum elevation of that ball and this is equal to the initial speed B not of the box is zero. So this term on the right hand side, the first term goes to zero, we're left with one half multiplied by negative 9. m per second squared, multiplied by the time. OK? And the time that the box collide with the ball is the same for the ball in the box because they start simultaneously. So this is 0.612245 seconds squared. Simplifying on the right hand side, on the left hand side, we still have a negative In Brackets D -1.8367 m. And on the right hand side, we have negative 1.8, m. You can divide by the negatives and then add 1.8, 36, 7 m to both sides. And we get That the distance d should be 3.67, 34 m. So in order for the box to collide with the ball, when the ball is at its maximum elevation, we first figured out what that maximum elevation was. We figured out the time that it took to reach the maximum elevation. And then we looked at the box displacement in order to sort out that the box met the ball at the same time at that maximum elevation. Comparing this to our answer choices, we can see they are rounded to three significant digits And we found the position the student above the floor should be 3.67 m which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. c. For what value of h does the collision occur at the instant when the first ball is at its highest point?

Verified Solution

Key Concepts

Projectile Motion

Free Fall

Kinematic Equations