Asked by Renay

A ball is projected upward at time t= 0.0s, from a point on the roof 10m above the ground. The ball rises, then falls until it strikes the ground. The initial velocity of the ball is 58.5 m/s. At time t=5.97s, what is the approximate velocity of the ball? Neglect air resistance.
Answered by drwls
Solve this equation using t = 5.97:

V = Vo - g t = 58.5 - 9.8 t

That will be zero, indicating that the ball is at its highest elevation.
Answered by Renay
Why'd you use 9.8?
Answered by drwls
Because the acceleration of gravity (g) is 9.8 m/s^2.
Answered by Renay
Oh, okay, thanks.
Answered by l;lpkj
jip-;
Answered by sk8er
how'd you get zero?
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