A basketball is thrown straight down and bounces back up

1. Dec 19, 2015

BensonCa

1. The problem statement, all variables and given/known data
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?

3. 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.

2. Dec 19, 2015

haruspex

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

3. Dec 19, 2015

BensonCa

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.

4. Dec 19, 2015

haruspex

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.

5. Dec 20, 2015

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

6. Dec 20, 2015

BensonCa

Thank you for your responses and assisting me in the right direction, I really appreciate it

7. Dec 20, 2015

haruspex

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?

8. Dec 20, 2015

BensonCa

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....

9. Dec 20, 2015

haruspex

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?

10. Dec 20, 2015

BensonCa

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.

11. Dec 20, 2015

haruspex

You seem to have multiplied mass by impulse, not by velocity change.

12. Dec 20, 2015

BensonCa

m* change in velocity= 4.43 I= 8.06 then I tried 8.06/4.43= 1.819/ 1.8/ 2

13. Dec 20, 2015

haruspex

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.

14. Dec 20, 2015

BensonCa

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

15. Dec 20, 2015

haruspex

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.

16. Dec 20, 2015

BensonCa

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

17. Dec 20, 2015

haruspex

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.
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?
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.

18. Dec 21, 2015

BensonCa

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.