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

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 horizontal, `theta=45^(@)`. Then
`R_(max)=(u^(2))/g`
Here, `R_(max)=100m`
`:. (u^(2))/g=100m`
As `v^(2)-u^(2)=2as`
Here, v=0 (At highest point velocity is zero)
`a=-g, sH uarr+ve`
`:. H=(u^(2))/(2g)=100/2=50` (using (i))