Arock is thrown upward with a velocity of 18 meters per second from the top of a 38 meter high cliff, and it misses the

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.

frequency or wavelenth of em radiation in a vacuum

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right