[111] Throw-up problem - Launch/freefall

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AI Thread Summary
The discussion focuses on solving a physics problem involving a stone launched from a height of 1.4 meters with an initial speed of 19.8 m/s. The key equations for determining the maximum height and time until the stone hits the ground are provided, specifically D=Vot + (1/2)at^2 and V=at + Vo. The challenge arises from integrating the initial height into the calculations, with attempts yielding incorrect time values. A suggested approach includes using the formula y=(v^2-v0^2)/(2a) for height and adjusting the time calculation to account for both ascent and descent. The conversation emphasizes the importance of correctly incorporating the initial height into the equations for accurate results.
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Homework Statement


A stone is launched straight up by a slingshot. Its initial speed is 19.8 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.
(a) How high above the ground does the stone rise?
wrong check mark m
(b) How much time elapses before the stone hits the ground?

Homework Equations


D= Vot + (1/2)at^2
V=at+Vo


The Attempt at a Solution


Vi=19.8m/s
d=1.4m
a=-9.8m/s

I actually solved another question I had while typing it up on here.. maybe this one will be the same.

Ok, I've attacked this thing every way I can think of, and my problem is the integration of it being launched from 1.4m above the ground. I tried 1.4=19.8t+ (1/2)(-9.8)t^2, solving quadratically for t=3.9688 and t=.071987, neither of which are correct. I know finding the time is the key to the solution, but I am not sure how to integrate the 1.4m into the formula.

Can someone point me in the right direction?
 
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Let me rewrite this your formula for part A is y= (v^2-v0^2)/(2a).

After you get your y component your formula for Par B is: t= v-v0/a.

t = time
a = acceleration
y = distance (vertical)

And that should give you the answer for the problem.

Sorry went through 3 or so edits kept copying the wrong equation, that should be it thou. Times the answer by 2 to get the up time and downtime for Part B.
 
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