Question

A baseball is popped straight up into the air and has a hanging -time of 6 second .Determine the maximum height to which the ball rises?

Answer 1

Time taken to reach the maximum height = 6/2 = 3 s

Final velocity, v = 0

Using first equation of motion, v = u + at

0 = u - 10 × 3

u = 30 m/s

Now,

s = ut + 1/2(at)^2

s= 30 × 3 – 12 × 10 × 3 × 3

Hence, s = 90 – 45 = 45 m