Answer to Question #221794 in Physics for Zoe

Question #221794
A stone is vertically thrown upward with a speed of 13 m/s from the edge of a cliff 60m high.how much later does it reach the bottom of the cliff
1
Expert's answer
2021-08-02T08:41:02-0400

The kinematic equation of motion gives the following (y-axis directed upward, origin is at ground):


y=y0+v0tgt22y = y_0 + v_0t - \dfrac{gt^2}{2}

where y0=60my_0 = 60m is the inital coordinate (cliff height), v0=13m/sv_0 = 13m/s is the inital speed, g=9.81m/s2g = 9.81m/s^2 is the gravitational acceleration, and yy is the coordinate at time tt.

At the bottom of the cliff y=0y = 0, this, substituting it, obtain:


0=60+13t4.905t24.905t213t60=00 = 60 + 13t - 4.905t^2\\ 4.905t^2 - 13t - 60 = 0

Solving the quadratic equation and taking the positive solution, obtain:


t7.7st \approx 7.7s

Answer. 7.7 s.


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