A projectile is fired from the surface of the earth with a velocity of $5 ext{ m/s}$ and angle $ heta$ with

Question

A projectile is fired from the surface of the earth with a velocity of $5 	ext{ m/s}$ and angle $	heta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 	ext{ m/s}$ 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 $	ext{m/s}^2$) is: [Given, $g = 9.8 	ext{ m/s}^2$]

Accepted Answer

1. **Equation of Trajectory:** The path of a projectile is given by the equation: 
   $$ y = x 	an 	heta - \frac{g x^2}{2 v^2 \cos^2 	heta} $$
   where $y$ and $x$ are coordinates, $v$ is the initial velocity, $	heta$ is the angle of projection, and $g$ is the acceleration due to gravity .
2. **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 $	an 	heta$ is already the same since the angle $	heta$ is given as equal.
   - The term $\frac{g}{2 v^2 \cos^2 	heta}$ must also be equal for both cases.
3. **Derivation:** Since $	heta$ is constant, $\cos^2 	heta$ is constant. Therefore, the ratio $\frac{g}{v^2}$ must be constant:
   $$ \frac{g_{earth}}{v_{earth}^2} = \frac{g_{planet}}{v_{planet}^2} $$
4. **Calculation:**
   - Earth: $g_e = 9.8 	ext{ m/s}^2$, $v_e = 5 	ext{ m/s}$
   - Planet: $g_p = ?$, $v_p = 3 	ext{ m/s}$
   $$ g_p = g_e \left( \frac{v_p}{v_e} ight)^2 $$
   $$ g_p = 9.8 	imes \left( \frac{3}{5} ight)^2 = 9.8 	imes \frac{9}{25} = 9.8 	imes 0.36 $$
   $$ g_p = 3.528 \approx 3.5 	ext{ m/s}^2 $$

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