Ball dropped out a truck window, find bounce height and dist

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Homework Help Overview

The problem involves a ball dropped from a moving pickup truck and requires determining the height of the ball's first bounce and the horizontal distance it travels before the next bounce. The context includes concepts from mechanics, specifically dealing with energy conservation and the coefficient of restitution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the bounce height and the horizontal distance traveled after the bounce. There are attempts to clarify the relationship between the coefficient of restitution and the bounce height, as well as the method for calculating the time of flight and distance.

Discussion Status

The discussion includes various interpretations of how to calculate the bounce height and the horizontal distance. Some participants offer guidance on using the coefficient of restitution, while others question the assumptions made in the calculations. There is no explicit consensus on the correct approach yet, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are working with a specific height of 5 ft and a coefficient of restitution of 0.6. There is some uncertainty regarding the correct application of the coefficient in relation to energy and height calculations.

jbm1939
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Homework Statement


A ball weighing 0.2 lbs is dropped out the window of a pickup truck moving horizontally at 45 mph. If the C.O.R. (e) between the ball and the ground is equal to 0.6, find the height of the ball's first bounce and the horizontal distance it will travel before bouncing again.

I believe i have the height of the bounce, but am stuck on how far it will travel horizontally. I'll put below what i have so far.

The attempt at a solution:

mgh = 1/2mv2
Vf = Square rt(2gh)
square rt(2(32.2)(5ft)
=17.9 ft/s

V1 = 0.6(17.9ft/s)
= 10.76 ft/s

1/2mv2 = mgh
h=vi2 / 2g
=10.762 / 2(32.2)
=1.798 ft
 
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You can get the bounce height a bit more easily, it's just the original times the square of the coefficient.
What is the horizontal velocity after the bounce? What else do you need to figure out to find the distance to the next bounce?
 
Now you have found v1 right. From now on

1/2*g*t^2= v1 ===> with that you find the t value ( the time the ball will go upwards)

Then find 2t ( the time ball will bounce upwards and fall back down)

2t* 45 mph = the distance your ball travel before its second bounce. Hope it was helpful and i didnt write something wrong by mistake
 
Does anyone know if I found the first height correctly? Another student told me that you just take the orignal height times e to get that. That seems to simple to me but I could be wrong. So in this case it would be 0.6 x 5ft = 3ft
 
jbm1939 said:
Does anyone know if I found the first height correctly? Another student told me that you just take the orignal height times e to get that. That seems to simple to me but I could be wrong. So in this case it would be 0.6 x 5ft = 3ft
A quick trip to google indicates that the coefficient of restitution is a ratio of energy, not of velocity. Post #2 assumed the opposite.
Scratch that. I read too quickly. It is a ratio of speed.

If energy goes down by a factor of 0.6 then potential energy at the top of the bounce must go down by a factor of 0.6 and height at the top of the bounce will go down by a factor of... what?

Scratch that. Re-read post #2.
 

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