A metal ball of 20 ~g is projected at an angle 30^ with the horizontal with an initial velocity 10 ~ms ⁻¹ . If the mass and angle of projection are doubled keeping the initial velocity the same, the ratio of the maximum height attained in the former to the latter case is :

Step-by-step solution

The maximum height H attained by a projectile is given by the formula H = u^2 ^2 2g , where u is the initial velocity, is the angle of projection, and g is the acceleration due to gravity. In the first case, u₁ = 10 ms ⁻¹ and ₁ = 30^ . The maximum height is H₁ = u₁^2 ^2(30^ ) 2g = 10^2 (1/2)^2 2g = 100 1/4 2g = 25 2g . In the second case, the mass is doubled (which does not affect the height), the velocity remains u₂ = 10 ms ⁻¹ , and the angle is doubled, so ₂ = 2 30^ = 60^ . The maximum height is H₂ = u₂^2 ^2(60^