A cricketer can throw a ball to a maximum horizontal distance of 100m. To what height above the ground can the cricketer throw the same ball with...

It is well known that a maximum distance for a given initial speed is reached if the angle is `45` degrees above horizontal. Denote the initial speed as `V_0` and the corresponding horizontal distance as `D_(max).` Let's find `V_0.`

If thrown `45` degrees above horizontal, the horizontal distance is `D(t)=(V_0 t)/sqrt(2)` and the height is `H(t) = (V_0 t)/sqrt(2) - (g t^2)/2` (`1/sqrt(2)` is `cos(45)` and `sin(45)`).

A ball falls at `t_1gt0` when `H(t_1)=0,` i.e. `t_1 = 2/g*V_0/sqrt(2),` so `D(t_1)=V_0^2/(g)=D_(max).` Thus `V_0=sqrt(D_1 g).`

Great, now if a cricketer throws a ball vertically upward with the same initial speed, the maximum height of a ball would be `V_0^2/(2 g)=D_(max)/2.` The simplest way to obtain this formula is to consider the energy conservation law, `mgH=(m V_0^2)/2.`

So the maximum height is twice less than the maximum horizontal distance, i.e. it is 50 m.