Question

An astronaut on the Moon releases a rock from rest and allows it to drop straight downward. If the acceleration due to gravity on the Moon is 1.62m/s2 and the rock falls for 2.4s before hitting the ground, what is its speed just before it lands?

Answer 1

Answer:

speed before landing = 3.9 m/s    (3 s.f.)

Explanation:

As rock is released from rest, u = 0 m/s     a = 1.62 m/s²     t = 2.4 s     v = ?

                      v = u + at

                      v = 0 + (1.62 x 2.4)

                      v = 3.888 = 3.9 s    (3 s.f.)

Hope this helps!

Answer 2

The speed of the rock before hitting the ground is 3.89 m/s

The given parameters;

acceleration due to gravity on moon, g = 1.62 m/s²

time taken for the object to fall, t = 2.4 s

To find:

• the speed of the object before hitting the ground;

The maximum height of fall of the rock is calculated as;

$h = v_0t + \frac{1}{2} gt^2\\\\v_0 = 0\\\\h = \frac{1}{2} gt^2\\\\h = 0.5 \times 1.62 \times 2.4^2 \\\\h = 4.67 \ m$

The speed of the rock before hitting the ground is calculated as;

$v_f^2 = v_0 ^2 + 2gh\\\\v_f^2 = 0 + 2\times 1.62 \times 4.67\\\\v_f^2 = 15.13\\\\v_f = \sqrt{15.13} \\\\v_f = 3.89 \ m/s$

Thus, the speed of the rock before hitting the ground is 3.89 m/s

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