A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown an angle θ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1 x . The value of x is _____ .

Step-by-step solution

When the first ball of mass m is projected vertically upwards, the second ball of mass 2 m makes an angle θ with the vertical. The time of flight is the same for both the balls. Then, T 1 = T 2 ⇒ 2 u 1 g = 2 u 2 sin θ g ⇒ u 1 = u 2 sin θ         . . . 1 Maximum height of projectile is given by H = u 2 sin 2 θ 2 g Ratio of the heights is H 1 H 2 = u 1 2 u 2 2 sin 2 θ         . . . 2 From equation 1 and 2 , we have ⇒ H 1 H 2 = 1 . Thus