Alien Projectile

By conservation of energy on Earth

$v_E^2 = 2g_E h_E$

while on the mysterious Planet

$v_P^2 = 2g_P h_P$

We know $v_P = 3, v_E=5, g_E = 9.8$ and since the trajectories are identical $h_E = h_P$ . Dividing the equations we get

$\displaystyle{ \dfrac{v_P^2}{ v_E^2} = \dfrac{2g_P h_P}{2g_E h_E}}$

$\displaystyle{ \dfrac{9}{25} = \dfrac{2g_P h_E}{2\cdot 9.8\cdot h_E} =\dfrac{\not2 g_P \not h_E}{\not 2\cdot 9.8 \not h_E} }$

$\displaystyle{g_P=\dfrac{9 \cdot 9.8 }{25} = 3.5 \dfrac{\textrm{m}}{\textrm{s}^2}}$

Since, the trajectories of the two planets are the same, their equations as well as their respective angles of projections will be same. Let the acceleration due to gravity on the second planet be A

We know that the equation of a projectile is:

$y=xtan\theta -\large\frac{gx^2}{2u^2cos^2\theta}$

$\implies \large xtan\theta -\large\frac{gx^2}{2(5)^2cos^2\theta}=xtan\theta -\large\frac{Ax^2}{2(3)^2cos^2\theta}$

$\implies \large\frac{g}{50}=\large\frac{A}{18}$

$\implies A=3.528ms^{-2}$