
#1
Sep2007, 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. 



#2
Sep2007, 09:30 PM

Sci Advisor
HW Helper
P: 8,961

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




#3
Sep2007, 10:44 PM

P: 336




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