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A particle is projected at an angle of $30^{\circ}$ from the horizontal with a speed of $10 \mathrm{~m} / \mathrm{s}$ on the earth. It's time of flight, range, velocity of impact are respectively $\mathrm{T}, \mathrm{R}, \mathrm{V}$. If this projectile is projected on the moon then its time of flight, range and velocity of impact is (assume gravitational field on moon is $\mathrm{g} / 6 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$6 T, 6 R, V$
Given, angle of projection, $\theta=30^{\circ}$ Initial speed of particle, $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$
Time of flights, range and velocity of impact on the surface of earth are T, R and V respectively
$\therefore \quad \mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}, \mathrm{R}=\frac{\mathrm{v}^2 \sin 2 \theta}{\mathrm{g}}$
Gravitational acceleration on moon's surface is equal to $\frac{1}{6}$ part of gravitational acceleration on the earth's surface.
Hence,
$\begin{aligned}
& g_m=\frac{g_e}{6} \\
& T_m=\frac{2 v \sin \theta}{\frac{g}{6}}=\frac{6 \times 2 v \sin \theta}{g}=6 \mathrm{~T} \\
& R_m=\frac{v^2 \sin 2 \theta}{\frac{g}{6}}=\frac{6 v^2 \sin 2 \theta}{g}=6 R
\end{aligned}$
Velocity of impact will be same at both places of moon and earth.
$\therefore \quad \mathrm{V}_{\mathrm{m}}=\mathrm{V}$
Time of flights, range and velocity of impact on the surface of earth are T, R and V respectively
$\therefore \quad \mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}, \mathrm{R}=\frac{\mathrm{v}^2 \sin 2 \theta}{\mathrm{g}}$
Gravitational acceleration on moon's surface is equal to $\frac{1}{6}$ part of gravitational acceleration on the earth's surface.
Hence,
$\begin{aligned}
& g_m=\frac{g_e}{6} \\
& T_m=\frac{2 v \sin \theta}{\frac{g}{6}}=\frac{6 \times 2 v \sin \theta}{g}=6 \mathrm{~T} \\
& R_m=\frac{v^2 \sin 2 \theta}{\frac{g}{6}}=\frac{6 v^2 \sin 2 \theta}{g}=6 R
\end{aligned}$
Velocity of impact will be same at both places of moon and earth.
$\therefore \quad \mathrm{V}_{\mathrm{m}}=\mathrm{V}$
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