A cricketer can throw a ball to a maximum horizontal distance of 100 m. how much high above the

We know that horizontal range is maximum if the angle of projection is 45^{o} and is given by

R_{max} = u^{2} / g

Here, R_{max} = 100 m

so, u^{2} / g = 100 m

If the cricketer throws the ball vertically upwardthen the ball will attain the maximum height from the ground

Hmax = u^{2} / 2g = 100 /2 = 50 m.