Question

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 1

Step 1: Find velocity of projection using range of projectile motion.
Given, maximum horizontal distance (𝑅) = 100m
The ball will cover maximum horizontal distance, only when the cricketer throws the ball with angle of projection of $45{}^{\circ }$ i.e. $\theta ={45}^{\circ }$.
Now, horizontal range is given by $R=\frac{u^2\sin 2\theta }{g}$
Where u = initial velocity
$100=\frac{u^2\sin {90}^{\circ }}{g}$
$u^2=100g$
$u=10\sqrt{g}....\left(1\right)$

Step 2: Find maximum height attained by the ball.
The ball can attain maximum height when it is thrown vertically upwards.
At maximum height (H), final velocity of ball becomes zero.
Acceleration a = - g
using third equation of motion,
$v^2=u^2+2as$
$v^2-u^2=-2gH$
$H=\frac{(0{)}^2-(100g)}{-2g}$ (Using equation (1))
$H=50m$.

Final answer:
The cricketer can throw the same ball 50 m high above the ground.