(a) With what speed must a ball be thrown vertically up in or- der to rise to a maximum height of 53.7 m? (b) For how long will it be in the air?

Given:
Maximum height to which ball rises,x = 53.7m .
Final speed of ball,v = 0 m/s .
Acceleration due to gravity, g = 9.8m/s².
(a) From the equation of kinematics, we have


The ball decelerates while moving upward because the gravity acts on it. Therefore acceleration of the ball is equal to the magnitude of acceleration due to gravity, but it acts downward opposite to the direction of its initial velocity.
Thus one can conclude that the acceleration (a)of the ball is -g.
Substitute the given values in the equation above to obtain initial speed as:


Therefore the ball was thrown upward with initial speed of 32 m/s.
(b) Let t represents the time taken by the ball for which it stays in the air until it reaches the maximum height x.
From the equation of kinematics, we have
v = v₀ + at
Substitute the given values v, a and the calculated value of v₀ to obtain time t as:

Rounding off to two significant figures
t = 3.3 s
Therefore the time for which the ball stays in the air is 3.3 s.