Question

A cricketer can throw a ball to a maximum horizontal distance of 100m.How much high above the ground can he throw same ball

Answer 1

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