- #1
jfeyen
- 7
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Homework Statement
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
Homework Equations
acceleration of gravity: 9.80 m/s2 ; not sure if I need to use -9.8 instead.
the equations we've been given so far include:
v=v0+at
v2=v02+2a(x-x0)
x=x0+((v0+v)/2)t
x=x0+v0t+(1/2)at2
I'm having a hard time figuring out which formula(s) would be most helpful
The Attempt at a Solution
I think I have to find how high the rock went first so I can then subtract it from the total free-fall to the bottom of the cliff at the end to find the height of just the cliff, but I'm having difficulties knowing what to do first and where to plug things in.
If I use v2=v02+2a(x-x0) and plug in my initial velocity of 8 m/s and I'm guessing a final velocity of 0 m/s when it stops at the top before coming back down, and an acceleration of -9.8 m/s2? and an initial x of 0 I come up with a final x of 3.27 m.
Not sure if I'm on the right track at all but another place I thought about starting was using v=v0+at to find the time it took just to reach the top of the throw, right before the rock free-falls to the bottom. When I did that I put in 0 m/s for my final velocity = 8 m/s for my initial velocity + (-9.8) for the acceleration of gravity x my unknown of t and found the time for the rock to go up before starting to come back down to be .816 s. Meaning that the free-fall then took 2.35s-.816s= 1.53s.
This is where I'm now stuck. Not sure what to to with both/either of these starting points and where I'm going to be able to pinpoint just the height of the cliff. If someone could let me know if I'm on the right track at all or point me in the right direction I would be extremely grateful!