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.

- physics -
**ailabianca**, Friday, January 10, 2014 at 7:34pm
the other straight downward from the top of the cliff. The stones are thrown with the same speed. The height of the cliff is 6.00 m, and the speed with which the stones are thrown is 9.00 m/s. Find the location of the crossing point.

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