Aball thrown straight up climbs for 3.0 sec before falling. neglecting air resistance, with what velocity was the ball thrown?

Speed, u = 29.4 m/s

Explanation:

Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s

Let u is speed with which the ball is thrown up. When the ball falls, v = 0

Using first equation of motion as :

v = u + at

Here, a = -g

So, u = g × t

u = 29.4 m/s

So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.

the answer would be c