Asked by Confused
A tennis ball is dropped from 1.80 m above the ground. It rebounds to a height of 1.00 m (assume down to be negative). (a) With what velocity does it hit the ground? (b) With what velocity does it leave the ground? (c) If the tennis ball were in contact with the ground for 0.030 s, find its acceleration while touching the ground.
Answers
Answered by
Henry
a. V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*1.8 = 35.28
V = 5.94 m/s.
b. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*1m = 19.6
V = 4.43 m/s. = Velocity with which it
hits gnd. after it bounces. = Velocity
with which it leaves gnd after it bounces.
V^2 = 0 + 19.6*1.8 = 35.28
V = 5.94 m/s.
b. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*1m = 19.6
V = 4.43 m/s. = Velocity with which it
hits gnd. after it bounces. = Velocity
with which it leaves gnd after it bounces.
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