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

Explanation

`R_max=100m=v^2/g`As` v^2-u^2=2as`
Here `v=0, a=-g s=R_max=100m`
`therefore(0^2)-u^2=2(-g)s or S=1/2u^2/g`
Since u=v, S=1/2 v^2/g = 1/2xx100m=50m`