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Ber [7]
3 years ago
6

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

nd hits the ground 1.37 s later. Part A) What is the acceleration due to gravity on this asteroid?
Physics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer:

a_y=0.92m/s^2

Explanation:

To solve this problem we use the formula for accelerated motion:

y=y_0+v_{y0}t+\frac{a_yt^2}{2}

We will take the initial position as our reference (y_0=0m) and the downward direction as positive. Since the rock departs from rest we have:

y=\frac{a_yt^2}{2}

Which means our acceleration would be:

a_y=\frac{2y}{t^2}

Using our values:

a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2

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Taya2010 [7]

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