Question

A rock is thrown upward with a velocity of 18 meters per second from the top of a 38 meter high cliff, and it misses the cliff on the way back down. when will the rock be 12 meters from the water, below? round your answer to two decimal places.

Answer 1

The rock would be at a point 12 m from water at a time 4.8 s.

Take the origin of the coordinate system at the top of the cliff. It is thrown upwards with a velocity u. When the rock is at a point 12 m from water, calculate the vertical displacement of the rock from the origin.

$y= -38m- (-12 m)=-26 m$

Use the equation of motion,

$y=ut +\frac{1}{2} at^2$

The rock falls under the acceleration due to gravity, directed down wards.

Substitute 18 m/s for u, -26 m for y and -9.8 m/s² for a=g.

$y=ut +\frac{1}{2} at^2\\ (-26 m)=(18m/s)t+\frac{1}{2}(-9.8m/s^2)t^2$

Solve the quadratic equation for t.

$t=\frac{(18m/s)(+/-)\sqrt{(18m/s)^2-4(4.9m/s^2)(-26m)} }{2(4.9m/s^2)}$

Taking only the positive value,

$t=\frac{(18 m/s)+(28.87 m/s)}{9.8 m/s^2)} \\ t=4.78 s$

After a time of 4.8 s the rock would be at a distance of 12 m from water.