Question

O. A player throws a ball at an initial velocity
of 36 m/s the main distance the ball can
reach (assume ball is caught at the same height
at which it was released) is​

Answer 1

Answer:

64.8 m

Explanation:

Let max. height be h m.

acceleration = -g = -10 m/s^2

final velocity (v) =0 m/s

Initial velocity (u) = 36 m/s

v^2 = u^2 + 2gh

=> 0 = 36^2 - 20h

=> h = 1296/20 = 64.8