Question

A ball is taken to Mars where the acceleration due to gravity is approximately $4m{s}^{-2}\left(0.4g_e\right)$. It is thrown from ground level with the initial velocity $v_0$, in the upward direction. On Mars, it will be in the air for ____ seconds and reach a maximum height _____.t and y are time of flight and maximum height when the ball is thrown on earth.

• A. 0.4 t, 2.5 y
• B. 0.4 t, 6.25 y
• C. 2.5 t, 2.5 y
• D. 6.25 t, 2.5 y

Answer 1

The correct option is C 2.5 t, 2.5 y

Velocity with which ball is thrown is same, upto topmost point.

$\frac{{v_m}^2}{{v_e}^2}=1=\frac{2\times 0.4\times g_e\times H_m}{2\times g_e\times H_e}\therefore H_m=2.5H_e\Rightarrow 2.5ys=0\times t+\frac{1}{2}g{t}^2\frac{H_m}{H_e}=\frac{\frac{1}{2}\times 0.4\times g_e\times {t_m}^2}{\frac{1}{2}\times g_e\times {t_e}^2}\therefore {t_e}^2\times 6.25={t_m}^2\Rightarrow t_m=2.5t$