Projectile motion
Classical Mechanics
Level
2
A cricketer can throw a ball to a maximum horizontal distance of . To what height above the ground can the cricketer throw the same ball with same speed?
The answer is 50.
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1 solution
d = v^2 * sin(2 * angle) / g For maximum distance, the angle is 45 degrees. So, 100 = v^2 /g => v^2 = 100g ------ (1)
Now, for maximum height, velocity along horizontal axis would be zero and so we can apply this formula 2as = v^2- u^2 => -2gs = 0 - 100 g -----> From (1)
=> s = 100g/(2g) => so, s = 50 meters