Aplayer holds two baseballs a height (h) above the ground. he throws one vertically upward at speed (vo) and the other one vertically downward at the same speed. determine the speed of each ball when it strikes the ground and the difference between their times of flight.

Assume that air resistance may be ignored.
All measurements are positive upward.
g =  acceleration due to gravity.

Case A; The ball is thrown vertically upward.
The time, t₁, to reach maximum height is one half of the time of flight.
Because the vertical velocity is zero at maximum height,
V₀ - gt₁ = 0
t₁ = V₀/g.
The time of flight is
t = 2t₁ = (2V₀)/g.                      (1)
Let v =  the vertical velocity with which the ball strikes the ground. Then
v² = V₀² + 2(-g)(-h)
v = √(V₀² + 2gh)                      (2)

Case B: The ball is thrown vertically downward.
The time of flight, t, is given by
-h = -V₀t + (-g)t²
gt² + V₀t - h = 0
t = 1/(2g)[-V₀ +/- √(V₀² + 4gh)]
Reject negative time
   (3)

Let v= the speed with which the ball strikes the ground.
-v = -V₀ - gt
v = V₀ + gt


The speed with which the ball strikes the ground is
v = V₀√[1 + (2gh)/V₀²], when the ball is thrown upward
when the ball is thrown downward.

The difference in time of flight is