An astronaut on the starship Enterprise is roaming around on a distant planet. He drops a rock from the top of a cliff and observes that it takes time t1 = \sqrt{2} sec to reach the bottom. He now throws another rock vertically upwards so that it reaches a height h = 10 m above the cliff before dropping down the cliff. The second rock takes a total time t2 = 2 sec to reach the bottom of the cliff, starting from the time it leaves the astronaut’s hand. The planet has a very thin atmosphere which offers negligible air resistance. What is the value of acceleration (in m/s2) due to gravity on this planet?

Engineering

Physics

Work Done by a Central Force

Question

An astronaut on the starship Enterprise is roaming around on a distant planet. He drops a rock from the top of a cliff and observes that it takes time t₁ = $\sqrt{2}$ sec to reach the bottom. He now throws another rock vertically upwards so that it reaches a height h = 10 m above the cliff before dropping down the cliff. The second rock takes a total time t₂ = 2 sec to reach the bottom of the cliff, starting from the time it leaves the astronaut’s hand. The planet has a very thin atmosphere which offers negligible air resistance. What is the value of acceleration (in m/s²) due to gravity on this planet?

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$v_{2} = \sqrt{2 g \times 10} = \sqrt{20 g}$ $H = \frac{1}{2} \times g \times 2$

$- H = 2 \sqrt{20 g} - \frac{1}{2} \times g \times 4$

$- g = 2 \sqrt{20 g} - 2 g \Rightarrow g = 2 \sqrt{20 g} \Rightarrow g^{2} = 80 g \Rightarrow g = 80$

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