Question

player throws a ball at initial velocity 36 m/s .what will be the maximum distance reached by the ball(assume ball is catched at the same point it was throwed)​

Answer 1

Explanation:

The distance is same as the ball covers during rise and distance covers during fall.

If we take distance covered be "s"

Initial velocity be "u"

Final velocity be "v"

Acceleration= -9.8 m/s (Because we oppose the force of gravitation)

At some point ball will be stopped and start falling .At this point final velocity=0

So, by using 2nd and 3rd eq. of motion you can easily solve this.

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

or, v^2=u^2+2as

Solve this by yourself. If I do all the work What are you gonna do!!