The acceleration due to gravity on a planet i | vedclass

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(D) To ensure safety,the velocity attained upon reaching the ground must be the same on both the Earth and the planet.Let $g_1 = 9.8 \, m/s^2$ be the

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$15$

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(D) To ensure safety,the velocity attained upon reaching the ground must be the same on both the Earth and the planet.Let $g_1 = 9.8 \, m/s^2$ be the acceleration due to gravity on Earth and $h_1 = 3 \, m$ be the safe height on Earth.Let $g_2 = 1.96 \, m/s^2$ be the acceleration due to gravity on the planet and $h_2$ be the corresponding safe height on the planet.The velocity $v$ attained after falling from height $h$ is given by $v = \sqrt{2gh}$.Equating the velocities for safety: $\sqrt{2 g_1 h_1} = \sqrt{2 g_2 h_2}$.Squaring both sides: $g_1 h_1 = g_2 h_2$.Substituting the values: $9.8 	imes 3 = 1.96 	imes h_2$.$h_2 = \frac{9.8 	imes 3}{1.96} = 5 	imes 3 = 15 \, m$.Thus,the corresponding height on the planet is $15 \, m$.

The acceleration due to gravity on a planet is $1.96 \, m/s^2$. If it is safe to jump from a height of $3 \, m$ on the Earth,the corresponding height on the planet will be ........ $m$.

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