A ball is projected upward at time t = 0.0 s, from a point on a roof 10 m 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 = 5.97 s, what is the approximate velocity of the ball? Neglect air resistance. +175 m/s -175 m/s 12 m/s zero -12 m/s

(1) Answers

First solve the equation using t=5.97 V= Vo - g t = 58.5 - 9.8t That will be zero so...indicate that the ball is at the highest elevation. The reason why I used 9.8 is because the acceleration of gravity (g) is 9.8 m/s^2.  hope it helped :) 

Add answer