Asked by Clement
A table tennis ball is dropped unto the floor from a height of 4m and it rebounds to a height of 3m. If the time of contact with the floor is 0.01secs, what is the magnitude and direction of the acceleration during the contact
Answers
Answered by
Henry
h = 0.5g*t^2 = 4 m.
4.9*t^2 = 4
t^2 = 0.816
Tf = 0.904 s. = Fall time.
V1^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4
V1 = 8.85 m/s., Downward.
V^2 = V2^2 + 2g*h = 0
V2^2 = -2g*h = -2*(-9.8)*3 = 58.8
V2 = 7.67 m/s, Upward.
a = (V2-V1)/t = (7.67-8.85)/0.01 = -118
m/s^2, Upward.
4.9*t^2 = 4
t^2 = 0.816
Tf = 0.904 s. = Fall time.
V1^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4
V1 = 8.85 m/s., Downward.
V^2 = V2^2 + 2g*h = 0
V2^2 = -2g*h = -2*(-9.8)*3 = 58.8
V2 = 7.67 m/s, Upward.
a = (V2-V1)/t = (7.67-8.85)/0.01 = -118
m/s^2, Upward.
Answered by
Henry
Note: The Fall time is not required.
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