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∘ i.e. θ=45∘.
Now, horizontal range is given by R=u2sin2θg
Where u = initial velocity
100=u2sin90∘g
u2=100g
u=10√g....(1)
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,
v2=u2+2as
v2−u2=−2gH
H=(0)2−(100g)−2g (Using equation (1))
H=50 m.
Final answer:
The cricketer can throw the same ball 50 m high above the ground.