Question

A body is projected upward making an angle a horizontal with velocity 300m p s find value of a so that horizontal range be maximum . Find range and time of flight

Answer 1

Answer:

a = 45°

Explanation:

For maximum range angle of projection should be 45°

R = u^2sin2a/g = (300)^2×sin90/g

= 90000×1/10 = 9000m or 9km

T = 2usina/g = 2×300×sin45/10

= 60×1/√2 = 60×0.707 = 42.42 sec