A boy drops a stone vertically down from the margin of a 24-m cliff. At the same

Question

A boy drops a stone vertically down from the margin of a 24-m cliff. At the same moment, his friend at the bottom of the cliff throws another stone vertically upward with an initial velocity of 12 m/s. After how many seconds will the two stones pass each other, given gravitational acceleration g=10m/s^2? Please provide an explanation step-wise and use significant figures.  I believe I have the solution correct but there is a previous chegg question asking the same thing but with two different 'correct' answers.

Explanation / Answer

from the top when stone is thrown initial velocity is zero.

Thus s = ut + 1/2*at^2

s = 1/2*10*t^2 = 5t^2

and from the bottom we have an equation

s1 = 12t - 5t^2

they meet when s1 + s = 24

thus

12t - 5t^2 +5t^2 = 24

t=2s

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