Homework Help: physics

Posted by Tracie on Wednesday, September 5, 2012 at 12:55am.

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 5.97 m. The stones are thrown with the same speed of 8.50 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

physics - Henry, Thursday, September 6, 2012 at 5:27pm
d1 + d2 = 5.97 m.
Vo1*t+o.5g*t^2 + Vo2*t+0.5g*t^2 = 5.97
8.5t + 4.9t^2 + 8.5t - 4.9t^2 = 5.97
17t = 5.97
t = 0.351 s. = Time at which they met.

h1 =5.97-(8.5*0.351 + 4.9*(0.3510^2) = 2.38 m. Above base of cliff.

h2 = 8.5*0.351 - 4.9(0.351)^2 = 2.38 m.
Above the base of cliff.

The stones cross paths at 2.38 m above base of cliff.

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