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

Let u be the velocity of projection of the ball. The ball will cover maximum horizotal distance when angle of projection with horizontal, `theta = 45^(@)`. Then , `R_("max") = u^(2)//g`
Here `u^(2)//g = 100 m` ……..(i)
In order to study the motion of the ball along vertical direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of Y-axis. Taking motion of the ball along vertical upward direction we have.
`u_(y) = u, a_(y) = -g, v_(y) = 0, t = ?, y_(0) = 0, y = ?`
AS `u_(y) = u_(y) + a_(y t)`
`therefore 0 = u + (-g)t` or `t = u//g`
Also `y = y_(0) + u_(y t) + (1)/(2) a_(y) t^(2)`
`therefore y = 0 + u(u//g) + (1)/(2)(-g)u^(2)//g^(2)`
`= (u^(2))/(g) - (1)/(2)(u^(2))/(g)`
`= (1)/(2) (u^(2))/(g) = (100)/(2) = 50 m` [from (i)]