Answer :
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
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 [tex]19.6m/s[/tex].
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 [tex]9.8m/s^{2}[/tex]. Notice the [tex]s^{2}[/tex] 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 [tex]2 * 9.8m/s[/tex] = [tex]19.6m/s[/tex] .
