### Q.16:- A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

**Answer:-**

Maximum horizontal distance,* R *= 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., *θ* = 45°.

The horizontal range for a projection velocity *v*, is given by the relation:

*R* = *u*^{2} Sin 2θ / g

100 = *u*^{2} Sin 90^{0} / *g*

*u*^{2} / *g* = 100 ….(i)

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity *v* is zero at the maximum height *H*.

Acceleration, *a = –g*

Using the third equation of motion:

*v*^{2} – *u*^{2} = -2*gH*

*H* = *u*^{2} / 2*g* = 100 / 2 = 50 m