A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?

Options

A. 50 m
B. 100 m
C. 150 m
D. 200 m

Answer

A

Explanation

Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizonatal, `theta=45^(@)" The "R_(max)=(u^(2))/(g)`
Here, `R_(max)=100m:.(u^(2))/(g)=100" "......(i)`
As `v^(2)=u^(2)=2as`
Here, v=0 (At highest point velocity is zero )
`a=-g ,s=H=:.H=(u^(2))/(2g)=(100)/(2)=50" "underset("Using"(i))uarr+v`