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?
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:
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 \(2 * 9.8m/s\) = \(19.6m/s\) .