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Question

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

a) $8.26 \textrm{ m}$

b) $0.717 \textrm{ s}$

Solution Video

# OpenStax College Physics Solution, Chapter 2, Problem 47 (Problems & Exercises) (6:50)

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Submitted by josemart015 on Tue, 07/23/2019 - 19:23

Why is acceleration still negative in part b?

Submitted by SayedAbdul on Thu, 07/23/2020 - 00:12

the time given is the entire time of the rock to go up and come down and pass the cliff and fall down to the ground. However, you have calculated the height of the cliff using the entire time (2.35s) instead of using the time at which it starts from the cliff to the ground. Pls correct me if I am wrong?

Submitted by ShaunDychko on Fri, 07/24/2020 - 14:35

Hi SayedAbdul, thanks for the question. You understand the given time correctly as the total time after launching the rock upward. It sounds to me like you're wishing for the time when the rock passes the top of the cliff going downward, and that you'd like to use this time to calculate the cliff height? This would be fine, but involves more work to find that time: You would have to calculate the time for a rock launched upward to return to "ground" (which is in quotes since in this case the rock would return to the top of the cliff, slightly displaced to the right so that it will go down the cliff), and then subtract this time from the total flight time.
As a shorter method, I'm using the total flight time in the displacement formula, taking the final position to be zero. The initial velocity is positive to reflect its upward direction, and acceleration is negative, and then $y_o$ is the initial position, namely the cliff height.
Hope this helps,
Shaun