# A basketball is thrown straight down and bounces back up

• BensonCa
In summary: I want to say that we can apply that the ball was at rest before it was thrown but since it doesn't specifically state that the ball is at rest I'm questing if we can assume that it was. That would mean that it gives the Velocity int however I don't see how that would help solve this problem. They also don't give the initial height either and with the forced applied via the ground and before the bounce I can assume that the Velocity afterwards would be greater than the initial velocity...No, that's not relevant. You know the velocity with which it hits the ground. What went before is not interesting.
BensonCa

## Homework Statement

A 0.624kg basketball is thrown straight down and strikes the floor at a speed of 7.10m/s. The floor applies an average force of 67.2N for a duration 0.120s, causing the ball to bounce back upward. How high does it bounce?

## The Attempt at a Solution

I'm not actually sure on the equations needed to figure out this problem, I guess that is why I'm having a hard time with it. I know were dealing with projectile motion but I also know it has to deal with peak height which is :
y = viy • t + 0.5 • g • t2

I think it's the bounce of the object and 0.120s of 67.2N force that has me thrown off. [/B]

The first step is to find the velocity immediately after bounce.
What equations do you know regarding momentum and impulses?

haruspex said:
The first step is to find the velocity immediately after bounce.
What equations do you know regarding momentum and impulses?
Impulse = Change in momentum

· The impulse experienced by an object is the force•time.

· The momentum change of an object is the mass•velocity change.

· The impulse equals the momentum change.

BensonCa said:
Impulse = Change in momentum

· The impulse experienced by an object is the force•time.

· The momentum change of an object is the mass•velocity change.

· The impulse equals the momentum change.
Right, so apply those here.
What is the momentum of the ball just before it hits the floor?
What will be its change in momentum?
Signs are important here.you need to choose which is the positive direction. The usual is to choose up as positive.

BensonCa
A 0.624kg basketball is thrown straight down and strikes the floor at a speed of 7.10m/s. The floor applies an average force of 67.2N for a duration 0.120s, causing the ball to bounce back upward. How high does it bounce?

ok F= 67.2N t = 120s I= change in M= 0.624 * (-7.10) since the initial direction is up, however after the bounce the direction of the ball is up creating a positive change in velocity

haruspex said:
Right, so apply those here.
What is the momentum of the ball just before it hits the floor?
What will be its change in momentum?
Signs are important here.you need to choose which is the positive direction. The usual is to choose up as positive.
Thank you for your responses and assisting me in the right direction, I really appreciate it

BensonCa said:
A 0.624kg basketball is thrown straight down and strikes the floor at a speed of 7.10m/s. The floor applies an average force of 67.2N for a duration 0.120s, causing the ball to bounce back upward. How high does it bounce?

ok F= 67.2N t = 120s I= change in M= 0.624 * (-7.10) since the initial direction is up, however after the bounce the direction of the ball is up creating a positive change in velocity
You need to use all three of the principles you quoted in post #3. (The time is 0.120s, not 120s.). What haven't you used? What given data haven't you used?

haruspex said:
You need to use all three of the principles you quoted in post #3. (The time is 0.120s, not 120s.). What haven't you used? What given data haven't you used?
I want to say that we can apply that the ball was at rest before it was thrown but since it doesn't specifically state that the ball is at rest I'm questing if we can assume that it was. That would mean that it gives the Velocity int however I don't see how that would help solve this problem. They also don't give the initial height either and with the forced applied via the ground and before the bounce I can assume that the Velocity afterwards would be greater than the initial velocity...

BensonCa said:
I want to say that we can apply that the ball was at rest before it was thrown but since it doesn't specifically state that the ball is at rest I'm questing if we can assume that it was. That would mean that it gives the Velocity int however I don't see how that would help solve this problem. They also don't give the initial height either and with the forced applied via the ground and before the bounce I can assume that the Velocity afterwards would be greater than the initial velocity...
No, that's not relevant. You know the velocity with which it hits the ground. What went before is not interesting.
Your first bullet in post #3 mentioned force and time. You are given a force and a time. What can you do with those?

force * time equals impulse = 67.2N * 0.120s = 8.064, then impulse = momentum change which is m*change in V = 0.624kg * 8.06= 5.03 which I believe is wrong.

BensonCa said:
m*change in V = 0.624kg * 8.06
You seem to have multiplied mass by impulse, not by velocity change.

haruspex said:
You seem to have multiplied mass by impulse, not by velocity change.
m* change in velocity= 4.43 I= 8.06 then I tried 8.06/4.43= 1.819/ 1.8/ 2

BensonCa said:
m* change in velocity= 4.43 I= 8.06 then I tried 8.06/4.43= 1.819/ 1.8/ 2
I have no idea what you are doing. Just writing numbers makes it really hard to follow. Where does 4.43 come from?
Work symbolically:
Use m for mass of ball,
vi for velocity just before landing,
vf for velocity just after bouncing,
t for time,
F for force.
You need to write two expressions for the momentum change. One is obtained from force and time (as you did in post #10), the other from the mass and change in velocity. Write that expression.
Setting the two expressions equal gives the equation you need.
Do not plug in any numbers until you have that equation.

haruspex said:
I have no idea what you are doing. Just writing numbers makes it really hard to follow. Where does 4.43 come from?
Work symbolically:
Use m for mass of ball,
vi for velocity just before landing,
vf for velocity just after bouncing,
t for time,
F for force.
You need to write two expressions for the momentum change. One is obtained from force and time (as you did in post #10), the other from the mass and change in velocity. Write that expression.
Setting the two expressions equal gives the equation you need.
Do not plug in any numbers until you have that equation.

change in velocity= Vf-Vi
where Vf = 15.164 (Vi + a 67.2 as force applied increase the acceleration of the ball * t 0.120s)

15.164/7.10 = 2.135 or 2.14
Change in m= m* change in V= 0.624kg * 2.14=1.33

BensonCa said:
change in velocity= Vf-Vi
where Vf = 15.164 (Vi + a 67.2 as force applied increase the acceleration of the ball * t 0.120s)

15.164/7.10 = 2.135 or 2.14
Change in m= m* change in V= 0.624kg * 2.14=1.33
You're doing it again. I cannot follow where you get some of these numbers from. Work symbolically, no numbers!
You've chosen to figure out the acceleration instead of using momentum. That's fine, but show how you are calculating that acceleration.

haruspex said:
You're doing it again. I cannot follow where you get some of these numbers from. Work symbolically, no numbers!
You've chosen to figure out the acceleration instead of using momentum. That's fine, but show how you are calculating that acceleration.
Ok so what I did there is use the uam equation Vf= Vi + at where I used the force applied to the ball 67.2 as acceleration because after the bounce the change in velocity came from the force the ground applied to the ball. Then I used the 0.120s for time, took the new velocity 15.164 and divided by Vi (7.10) to get the change in Velocity that I needed in order to caculate the change in momentum which= m* change in V. That together was 0.624kg*2.14=1.33

BensonCa said:
I used the force applied to the ball 67.2 as acceleration
Force is not acceleration. You cannot use one as though it is the other. What is the relationship between the two?
Also, you need to consider all forces acting on the ball during that time.
BensonCa said:
the uam equation Vf= Vi + at ... I used the 0.120s for time, took the new velocity 15.164
You need to be careful with signs, as I mentioned in post #4. Choose a direction, up or down, as positive. Based on your choice, which of Vi, Vf and the acceleration are positive?
BensonCa said:
divided by Vi (7.10) to get the change in Velocity
If you are driving at 50kph then accelerate to 60kph, what is your change in velocity?
Anyway, you don't need to do this. Once you have Vf you can proceed to calculating the bounce height.

Force equals mass times acceleration and are directly proportional, in increase to force on an object means an increase in acceleration.
Vi is negative since the 7.10 is in a downward position and Vf is positive since it is in effect after the bounce of the ball and headed back in the upward direction.
change in velocity of 50kph and 60kph is 10kph. so change in velocity is the difference between the Vf and Vi.

## What is the force acting on the basketball when it is thrown down?

The force acting on the basketball when it is thrown down is the force of gravity. This force pulls the basketball towards the ground at a constant rate of 9.8 meters per second squared.

## Why does the basketball bounce back up after hitting the ground?

The basketball bounces back up after hitting the ground due to the conservation of energy. When the basketball hits the ground, it deforms and stores potential energy. As it bounces back up, this potential energy is converted back into kinetic energy, causing the basketball to bounce back up.

## Does the mass of the basketball affect its bounce?

Yes, the mass of the basketball does affect its bounce. A heavier basketball will have more inertia and will bounce back up higher than a lighter basketball with the same force applied to it.

## How does the height from which the basketball is thrown affect its bounce?

The height from which the basketball is thrown affects its bounce because it determines the amount of potential energy the basketball has when it hits the ground. The higher the initial height, the more potential energy the basketball will have, resulting in a higher bounce.

## What other factors can affect the bounce of a basketball?

Other factors that can affect the bounce of a basketball include the surface it bounces on, the air pressure inside the basketball, and the angle at which it is thrown. These factors can all influence the amount of force and energy involved in the bounce, resulting in different heights and patterns of bounce.

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