Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

3.5

5.9

16.3

110.8

Step-by-Step Solution

Equation of Trajectory: The path of a projectile is given by the equation: $y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta}$ where $y$ and $x$ are coordinates, $v$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity .
Condition for Identical Trajectory: For two projectiles to follow the same trajectory (identical path $y(x)$ ), the coefficients of $x$ and $x^2$ in the equation must be identical.

The term $\tan \theta$ is already the same since the angle $\theta$ is given as equal.
The term $\frac{g}{2 v^2 \cos^2 \theta}$ must also be equal for both cases.

Derivation: Since $\theta$ is constant, $\cos^2 \theta$ is constant. Therefore, the ratio $\frac{g}{v^2}$ must be constant: $\frac{g_{earth}}{v_{earth}^2} = \frac{g_{planet}}{v_{planet}^2}$
Calculation:

Earth: $g_e = 9.8 \text{ m/s}^2$ , $v_e = 5 \text{ m/s}$
Planet: $g_p = ?$ , $v_p = 3 \text{ m/s}$ $g_p = g_e \left( \frac{v_p}{v_e} \right)^2$ $g_p = 9.8 \times \left( \frac{3}{5} \right)^2 = 9.8 \times \frac{9}{25} = 9.8 \times 0.36$ $g_p = 3.528 \approx 3.5 \text{ m/s}^2$

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