Answer to Question #135604 in Astronomy | Astrophysics for Joshua

Question

A ball is released from a height H and falls freely to the ground. It takes 3.0 s to reach the ground. Use g = 10 m/s2
What is the height of the ball above the ground after 2.0 seconds?

Answer 1

Let H be the height from where the ball is dropped. It takes 3sec to reach the ground. Using 3^rd kinematics equation. ,then,

s = ut +0.5gt²

H = 0 + (0.5)(10)(3²)

H = 45m

After 2sec, we have-

s = 0 + (0.5)(10)(2²)

s = 20m

Height of ball above the ground after 2sec = 45 - 20

= 25m