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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?

Sagot :

louiseee
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 [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] .

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