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. `150 m`
B. `250 m`
C. `350 m`
D. `50 m`

Answer

D

Explanation

`R_("max") =100m`
Since, `R_("max") =v^(2)/g rArr 100 = v^(2)/g`
using equation of motion
`v^(2)-u^(2)=2as`
Here, v=0, a`=-g`, s=`R_("max")= 100m`
`therefore` `(0)^(2)-u^(2)=2(-g) xx s`
`rArr s=1/2u^(2)/g`
Since, u=v
`rArr s=1/2(v^(2)/g) = 1/2 xx 100 = 50 m.`