Question

A projectile is fired from the surface of the earth with a velocity of $5\, m s^{- 1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3\,ms^{ - 1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $ms^{ - 2}$ ) is (Given $g= 9.8 \,ms^{ - 2}$)

• A. 3.5
• B. 5.9
• C. 16.3
• D. 110.8

Answer 1

The Correct Option is A

To solve this question, we need to understand the comparison of projectile motion on Earth and another planet. The key concept here is that the trajectory of a projectile depends on two parameters: the initial velocity and the gravitational acceleration where the projectile is fired.

Let us define the symbols:

• v_1 = 5\, m/s: Initial velocity of the projectile on Earth
• \theta: Angle of projection
• g_1 = 9.8\, m/s^2: Acceleration due to gravity on Earth
• v_2 = 3\, m/s: Initial velocity of the projectile on another planet
• g_2: Acceleration due to gravity on another planet

The formula for the range R of a projectile is given by:

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

Since both projectiles follow identical trajectories, their ranges must be equal:

\frac{v_1^2 \sin(2\theta)}{g_1} = \frac{v_2^2 \sin(2\theta)}{g_2}

The terms \sin(2\theta) cancel out from both sides:

\frac{v_1^2}{g_1} = \frac{v_2^2}{g_2}

Rearranging for g_2 gives:

g_2 = \frac{v_2^2 \cdot g_1}{v_1^2}

Substituting the values:

g_2 = \frac{(3\, m/s)^2 \cdot 9.8\, m/s^2}{(5\, m/s)^2}

= \frac{9 \cdot 9.8}{25}

= \frac{88.2}{25}

= 3.528\, m/s^2

Therefore, the acceleration due to gravity on the other planet is approximately 3.5\, m/s^2.

The correct answer is 3.5.