Standing side by side, you and a friend step off a bridge at different times and fall for 1.41 s to the water below. Your friend goes first, and you follow after she has dropped a distance of 1.93 m. (a)How long does it take your friend to fall 1.93 m? (b)What is the separation between the two of you when your friend hits the water?

asked by Kasey

11 years ago

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1 answer

t = √(2s/a) = √(2*1.93/9.8)
separation is thus 1/2 a(1.41^2 - (1.41-t)^2)

answered by Steve

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Question

Standing side by side, you and a friend step off a bridge at different times and fall for 1.41 s to the water below. Your friend goes first, and you follow after she has dropped a distance of 1.93 m. (a)How long does it take your friend to fall 1.93 m? (b)What is the separation between the two of you when your friend hits the water?

1 answer

t = √(2s/a) = √(2*1.93/9.8)
separation is thus 1/2 a(1.41^2 - (1.41-t)^2)

Ask a New Question or answer this question.

Steve · Answer

t = √(2s/a) = √(2*1.93/9.8) separation is thus 1/2 a(1.41^2 - (1.41-t)^2)