Question

An astronaut standing on a platform on the moon drops a hammer. if the hammer falls 6.0 meters vertically in 2.7 seconds, what is its acceleration?

Answer 1

The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,

    d = Vot + 0.5at²

where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,

   6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²

Determining the value of a,
    a = 1.646 m/s²

ANSWER: 1.646 m/s²

Answer 2

The acceleration of the hammer on the moon is 1.65 m/s²

The given parameters;

• height of fall, h = 6.0 m
• time of motion, t = 2.7 s

The acceleration of the hammer is calculated by applying the second equation of motion;

h = ut + ¹/₂at²

where;

• u is the initial velocity of the hammer = 0

6 = 0 + (0.5 x 2.7²) a

6 = 3.645a

$a = \frac{6}{3.645} \\\\a = 1.65 \ m/s^2$

Thus, the acceleration of the hammer on the moon is 1.65 m/s²

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