Question

Two projectiles are fired from the two imaginary planets A and B. The projectile is fired from the planet A with a velocity of $10m/s$ and at angle $\theta$ with horizontal. Another projectile is fired from the surface of planet B with a velocity of $6m/s$ at the same angle follows trajectory which is identical with the trajectory of projectile from planet A. What is the ratio of the acceleration due to gravity of planet A to planet B?

• A. $\frac{5}{3}$
• B. $\frac{3}{5}$
• C. $\frac{25}{9}$
• D. $\frac{9}{25}$

Answer 1

The correct option is C $\frac{25}{9}$

Given that,
In planet A,
velocity of projectile $\left(V_A\right)=10m/s$
Angle of projection $=\theta$
In planet B,
velocity of projectile $\left(V_B\right)=6m/s$
Let the acceleration due to gravity in planet A is $g_A$ and planet B is $g_B$.
They both have identical trajectory means their maximum height must be equal.
So, maximum height achieaved on planet A=maximum height achieved on planet B.
$\Rightarrow H_A=H_B$
$\Rightarrow \frac{v_A^{2}si{n}^2\theta }{2g_A}=\frac{v_B^{2}si{n}^2\theta }{2g_B}$
$\Rightarrow \frac{v_A^2}{g_A}=\frac{v_B^2}{g_B}$
$\Rightarrow \frac{g_A}{g_B}=\frac{v_A^2}{v_B^2}=\frac{{10}^2}{6^2}=\frac{25}{9}$