A projectile thrown with a speed v at an angle θ has a range R on the surface of earth. For same v and θ, its range on the surface of moon will be

The correct option is 6R.

Let's understand the physics behind this step-by-step.

Step 1: Formula for the Horizontal Range of a Projectile
When a projectile is launched from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal, its horizontal range $R$ is given by the formula:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

where $g$ is the acceleration due to gravity on the surface where the projectile is thrown.

Step 2: Compare Gravity on Earth and the Moon
The acceleration due to gravity on the surface of the Moon ($g_m$) is approximately one-sixth of the acceleration due to gravity on the surface of Earth ($g_e$):

$g_m=\frac{g_e}{6}$

Step 3: Calculate the Range on the Moon
Since the initial velocity $v$ and the launch angle $\theta$ remain the same on both Earth and the Moon, the range is inversely proportional to gravity:

$R\propto \frac{1}{g}$

Let $R_e=R$ be the range on Earth and $R_m$ be the range on the Moon. We can write the ratio of the ranges as:

$\frac{R_m}{R_e}=\frac{g_e}{g_m}$

Substitute $g_m=\frac{g_e}{6}$ into the equation:

$\frac{R_m}{R}=\frac{g_e}{(g_e/6)}=6$

Therefore, solving for $R_m$:

$R_m=6R$

Due to the weaker gravitational pull on the Moon, the projectile will stay in the air longer, resulting in a range that is 6 times larger than its range on Earth.