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.)

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

[tex]g=G\frac{M}{r^2}[/tex]

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

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.1\times 0.46+\frac{1}{2}\times 9.81\times 0.46^2\\\Rightarrow s=5.22\ m[/tex]

Distance travelled in 0.46 seconds is 5.22 m