What is the height of the cliff?

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Homework Help Overview

The problem involves two stones dropped from a cliff, with one stone being dropped first and the second thrown downward after a delay. The objective is to determine the height of the cliff based on their simultaneous impact with the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to analyze the motion of both stones. There is an exploration of whether the final velocities of both stones can be assumed to be the same upon impact. Questions arise about the relevance of calculating the distance traveled by the first stone before the second stone is thrown.

Discussion Status

Some participants have provided guidance on focusing on the height of the cliff and suggested using the equations directly related to the stones' motions. There is acknowledgment of the confusion regarding the timing of the second stone's throw, with one participant expressing a breakthrough in understanding the time difference between the two stones.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can provide. There is a noted lack of consensus on the best approach to take after calculating the speed of the first stone.

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Homework Statement



A stone is dropped from a cliff; 2.18 s later another stone is thrown downward with an initial speed of 31 m/s. They reach the ground simultaneously. Find the height of the cliff.

Homework Equations



vf^2 = vo^2 + 2ax
x = volt + 0.5at^2
vf = vo + at

The Attempt at a Solution



i've been working on this for the past two hours, and it's driving me nutsss.

i found the speed of the stone in 2.18 seconds by using the equation vf = vo + at. (t = 2.18 seconds; a = 9.81 m/s^2; vo = 0) i found the velocity to be 21.3858 m/s.

with that information, i found how far the stone traveled with the equation vf^2 = vo^2 + 2ax. (vf^2 = (21.3858)^2; vo^2 = 0; a = 9.81) the answer: 23.310522 m.

after that, i don't know what else to do with stone b. I've been playing around with the first equation, since the two stones hit the ground at the same time. is it wrong to assume that the final velocities of the two would be the same?

i would appreciate it very much if someone could throw me a bone on what to do with the second stone because I'm at a loss.

thanks!
 
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You have the right equations, and are plugging in values correctly. But you seem to be calculating unnececessary things.

Is it in any way useful to calculate how far the first stone has traveled after 2.18s?

Focus on thinking about what you really want, i.e. the height of the cliff. Think of x as being the height of the cliff. Then plug in values for your equations for the first stone, and then the second stone.

Given both sets of equations, solve for x.
 
i'm still stuck on what i need to do... i don't know where to go after finding the speed of the first stone at 2.18 seconds. i know a = 9.81 m/s^2, speed of the first stone is 21.3858 m/s when the second stone is thrown with an initial velocity of 31 m/s. after that, everything is coming up blank. I've tried to set equations equal to each other, but they always cancel out the wanted variable.
 
AAAHHH I GOT IT I GOT IT! i forgot that the time for the stone b is 2.18 seconds less than stone a! yayyy!
 

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