Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
