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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?
Solution
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 = (v^2sin2theta)/g`
`100 = v^2/g sin 90^@`
`v^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 = 2gH`
= `(0)^2 - (sqrt(100)g)^2 = 2(-g)H`
`H=(100g)/(2g)=50 "m"`
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