a. Between the time the ball is first dropped and the first time it hits the ground, the ball's vertical velocity v at time t is given by
v = -gt
and its height y is
y = 2.00 m - 1/2 gt²
Solve the second equation for time t when the height is y = 0 (corresponding to the moment the ball first hits the ground) :
0 = 2.00 m - 1/2 gt²
1/2 gt² = 2.00 m
gt² = 4.00 m
t² ≈ 0.408163 s²
t ≈ 0.638877 s ≈ 0.639 s
Solve for v in the first equation at time this t :
v = -gt
v ≈ -6.26099 m/s ≈ -6.26 m/s
(That is, the ball hits the ground with a downward velocity of 6.26 m/s.)
b. The ball loses some energy after hitting the ground, and when it bounces back up, so it does so with a smaller upward initial velocity v' ("v prime").
At its maximum height h, the ball has zero vertical velocity. It's in freefall, so acceleration is constant. Then
0² - (v')² = -2gh
Solve for v' using the fact that h = 1.10 m after the first bounce:
(v')² = 2g (1.10 m)
(v')² = 21.56 m²/s²
v' ≈ 4.64327 m/s ≈ 4.64 m/s
