Question

A ball is thrown straight up from the ground with an unknown velocity. It returns to the ground after 4.0seconds. With what velocity did it leave the ground?

Answer 1

V = 1/2at^2

a = 9.8 m/2^2 (constant)
t = 4.0 s

1/2 • 9.8 • 4^2
1/2 • 9.8 • 16

= 78.4 m/s

Answer 2

Answer:

The ball was thrown at approximately $19.6m/s$.

Explanation:

Assuming that there is no air resistance (or drag) in this question, the time to go up is the same to go down. Therefore, we can assume that it takes 2 seconds to go up and then 2 seconds to go down (4 seconds in total).

On Earth's surface, gravity is approximately $9.8m/s^{2}$. Notice the $s^{2}$ meaning that gravity is given in terms of acceleration. As it's an acceleration, it means that velocity changes at a rate of 9.8 meters per second. As we have 2 seconds of deceleration going up, we have a starting speed of  $2 * 9.8m/s$ = $19.6m/s$ .