From the top of a cliff, a person throws a stone straight downward. The initial

Q: From the top of a cliff, a person throws a stone straight downward. The initial

a) acceleration on the stone is acceleration due to gravity Its magnitude is 9.8 m/s^2 Its direction is downward Answer: 9.8 m/s^2 downward b) Taking downward direction as positive: vi = 9.7 m/s a = 9.8 m/s^2 t = 0.54 s use: d = vi*t + 0.5*a*t^2 = 9.7*0.54 + 0.5*9.8*(0.54)^2 = 5.24 + 1.43 = 6.67 m Answer: 6.7 m

ID: 1876905 • Letter: F

Question

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 9.7 m/s.

(a) What is the acceleration (magnitude and direction) of the stone while it moves downward, after leaving the person's hand?

Is the stone's speed increasing or decreasing?

increasingdecreasing

(b) After 0.54 s, how far beneath the top of the cliff is the stone? (Give just the distance fallen, that is, a magnitude.)
m

magnitude m/s2 direction ---Select--- upward downward The magnitude is zero.

Explanation / Answer

acceleration on the stone is acceleration due to gravity

Its magnitude is 9.8 m/s^2

Its direction is downward

Answer:

9.8 m/s^2

downward

Taking downward direction as positive:

vi = 9.7 m/s

a = 9.8 m/s^2

t = 0.54 s

use:

d = vi*t + 0.5*a*t^2

= 9.7*0.54 + 0.5*9.8*(0.54)^2

= 5.24 + 1.43

= 6.67 m

Answer: 6.7 m

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From the top of a cliff, a person throws a stone straight downward. The initial

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From the top of a cliff, a person throws a stone straight downward. The initial

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