Question

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

• A. $\frac{R}{6}$
• B. $6R$
• C. $\frac{R}{36}$
• D. $36R$

Answer 1

The correct option is (A) $6R$

Given,

Velocity = v

The angle of projection = $\theta$

Acceleration due to gravity = g

We know,

Range is given by $R=\frac{u^{2}sin2\theta }{g}$

We know that acceleration due to gravity is less on moon and it is one-sixth of the value on earth.

So,
On moon $g_m=\frac{g}{6}$

So range can be given as

$R_1=\frac{u^{2}sin2\theta }{\frac{g}{6}}=\frac{6u^{2}sin2\theta }{g}=6R$

Hence, the new range will be 6 times the previous range.

The correct option is (A)