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

Explanation

Let u be the velocity of projection
The maximum horizontal range , `R_("max")=(u^(2))/(g)`
`therefore (u^(2))/(g)=100`
The maximum height attrained ,
`=h(u^(2) sin ^(2) theta )/(2g)`
Foer the maximum vertical range , the angle of projection `theta=90^(@)`
Therefore ,`h=(u^(2) sin ^(2) 90^(@))/(2g) =(u^(2))/(2g)`
From (1) and (2) , we can write
`H_("max")=(1)/(2) R =(1)/(2) xx100 m=50m`