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:
The cricketer can throw the ball to a maximum horizontal distance of 100 m. To do this the ball has to be thrown with a velocity v at an angle of elevation equal to 45 degrees.
The range of an object thrown at a velocity v with an angle of elevation 45 degrees is S = v^2/g
100 = v^2/g
=> v^2 = 100*g
In the same throw, the maximum height of the ball can be derived from the equation v^2 - u^2 = 2*g*h
It is is h = =
= 50 metres
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