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Ball Freefall problem

by AraProdieur
Tags: freefall
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AraProdieur
#1
Sep20-07, 08:52 PM
P: 27
A ball is thrown vertically upward with an initial speed of 15 m/s. Then, 0.92 s later, a stone is thrown straight up ( from the same initial height as the ball) with an initial speed of 27 m/s. The acceleration of gravity is 9.8 m/s^2.
How far above the release point will the ball and stone pass each other? Answer in units of m.

So far, I have written down this equation 15( t + 0.92) - 4.9(t + 0.92)^2= 27t- 4.9t^2
And when I simplify everything, I get a t= .4593

Now, I don't know if that is even remotely correct, but if I am, I would like to know how to go about getting closer to the answer at hand.

Thank you.
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mgb_phys
#2
Sep20-07, 09:30 PM
Sci Advisor
HW Helper
P: 8,953
Sounds right, now just substitute t into one of the equations and get the height - better still do both equations just to check the answer
FedEx
#3
Sep20-07, 10:44 PM
P: 336
Quote Quote by AraProdieur View Post
A ball is thrown vertically upward with an initial speed of 15 m/s. Then, 0.92 s later, a stone is thrown straight up ( from the same initial height as the ball) with an initial speed of 27 m/s. The acceleration of gravity is 9.8 m/s^2.
How far above the release point will the ball and stone pass each other? Answer in units of m.

So far, I have written down this equation 15( t + 0.92) - 4.9(t + 0.92)^2= 27t- 4.9t^2
And when I simplify everything, I get a t= .4593

Now, I don't know if that is even remotely correct, but if I am, I would like to know how to go about getting closer to the answer at hand.

Thank you.
Its correct.As the displacement of both the oblects are same you can use the equality among the equations as you have done above.


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