Question

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 5.39 m. the stones are thrown with the same speed of 9.44 m/s. find the location (above the base of the cliff) of the point where the stones cross paths.

Answer 1

The sum of the speeds is 9.44 m/s + 9.44 m/s which is 18.88 m/s Note that the sum of the speeds of the two stones is always 18.88 m/s because as the upward moving stone loses speed, the downward moving stone gains the same amount of speed each unit of time. We can find the time for the stones to meet. t = d / v t = 5.39 m / 18.88 m/s t = 0.285487 seconds We can use the upward moving stone to find the height y. y = v0 t + (1/2) a t^2 y = (9.44 m/s)(0.285487 s) - (1/2) (9.8 m/s^2) (0.285487 s)^2 y = 2.30 m The two stones cross paths a height of 2.30 meters above the base of the cliff.