A tennis ball is dropped from 1.93 m above the ground. It rebounds to a height of 0.921 m.With what velocity does it hit the ground?The acceleration of gravity is 9.8 m/s2.(Letdown be negative.)Answer in units of m/s.

Answer 1

a = -9.8
v = Vinitial - 9.8 t = -9.8 t
z = Zinitial + Vinitial t - 4.9 t^2 = 1.93 - 4.9 t^2
at ground z = 0
so
4.9 t^2 = 1.93
t = sqrt (1.93/4.9)
so
v at ground = -9.8 sqrt (1.93/4.9)

Answer 2

alternatively
kinetic energy at ground = loss in potential energy
(1/2) m |v|^2 = m g h
|v| = sqrt (2 g h)
|v| = sqrt (2 * 9.8 *1.93)
= 6.15 down (negative)
same answer

Answer 3

V^2 = Vo^2 + 2g*h = 0 + 19.6*1.93 = 37.8,
V = 6.15 m/s.