Question:
A cricketer can throw a ball to a maximum horizontal distance of 100m.How much high above the ground cab 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 = (u2 Sin 2θ) / g
100 = (u2 Sin 900 )/ g
u2 / 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:
v2 – u2 = -2gH
H = u2 / 2g = 100 / 2 = 50 m
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