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.
[tex]y= -38m- (-12 m)=-26 m[/tex]
Use the equation of motion,
[tex]y=ut +\frac{1}{2} at^2[/tex]
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.
[tex]y=ut +\frac{1}{2} at^2\\ (-26 m)=(18m/s)t+\frac{1}{2}(-9.8m/s^2)t^2[/tex]
Solve the quadratic equation for t.
[tex]t=\frac{(18m/s)(+/-)\sqrt{(18m/s)^2-4(4.9m/s^2)(-26m)} }{2(4.9m/s^2)}[/tex]
Taking only the positive value,
[tex]t=\frac{(18 m/s)+(28.87 m/s)}{9.8 m/s^2)} \\ t=4.78 s[/tex]
After a time of 4.8 s the rock would be at a distance of 12 m from water.
