From the top of a cliff, a person throws a stone straight downward. the initial speed of the stone just after leaving the person's hand is 9.1 m/s. (a) what is the acceleration (magnitude and direction) of the stone while it moves downward, after leaving the person's hand? magnitude 9.81 correct: your answer is correct. m/s2 direction downward correct: your answer is correct. is the stone's speed increasing or decreasing? increasing decreasing correct: your answer is correct. (b) after 0.46 s, how far beneath the top of the cliff is the stone? (give just the distance fallen, that is, a magnitude.)

a) 9.81 m/s²

b) 5.22 m

Explanation:

Any object falling to the surface of Earth has a constant acceleration. It does not matter what the initial speed is the rate of change of velocity will be constant. The constant value is calculated by

g = Accleration due to gravity

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Earth = 5.972 × 10²⁴ kg

r = Radius of Earth = 6.371×10⁶ m

Plugging all the values we get

g = 9.81 m/s²

The speed of the stone is increasing. Downward direction is considered as positive.

Time = t = 0.46 s

u = Initial velocity = 9.1 m/s

s = Displacement

Equation of motion

Distance travelled in 0.46 seconds is 5.22 m

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