A ball of mass m is thrown vertically up. Another ball of mass 2 m is thrown at an angle with the vertical. Both of them stay in air for the same period of time. What is the ratio of the height attained by the two balls. 

For the ball thrown vertically upwards, the time taken by the ball to come back is:

T₁ = (2u₁) / g

For the ball projected at an angle θ with the vertical, the time of flight is:

T₂ = (2u₂ cosθ) / g

Since the time for both balls is the same,

(2u₁) / g = (2u₂ cosθ) / g   ⇒   u₁ = u₂ cosθ

Now,

h₁ = (u₁²) / (2g)   and   h₂ = (u₂² cos²θ) / (2g)

Hence,

h₁ / h₂ = (u₁²) / (u₂² cos²θ) = (u₂² cos²θ) / (u₂² cos²θ) = 1

Therefore, the heights attained by both balls are equal.