A ball is dropped from a height of 1.75m. It bounces on the floor and reaches a maximum height of 1.35m above the floor. The ball's weight is 98.5g.
a) What is the ball's velocity just before and just after the bounce?
b) How large is the impulse that the ball gets in the bounce?
c) Assume that the ball is in contact with the floor for 15 ms. What is the ball's average acceleration in the bounce?
a) mgh=mv2/2, where the heights are 1.75 and 1.35 respectively.
b) I = Δp =mv2-mv1
c) average acceleration = Δv/Δt
The Attempt at a Solution
Setting downward direction to negative, I've solved a) and b) in the following manner,
a) using the energy principles so that velocity before bounce, v1=(-5.863m/s)
and velocity after bounce, v2=5.149m/s
b) Taking the signs of the velocities into consideration: I=(0.985*5.15)-(0.985*-5.86)=0.985*[5.15-(-5.86)]≈10.8 kgm/s
c) I'm not sure whether Δv is 1)[5.863-5.149]
or 2) [5.149-(-5.863)],
because the first alternative gives an acceleration (setting Δt to 0.015s) of 47.6m/s2
And the second alternative gives a=734 m/s2.
In my opinion the second average acceleration seems excessively large, and so I'm unsure of which I should choose.. And furthermore, are the calculations in a) and b) correct? I could really use some help, thanks!