A stone is thrown at an angle of 30.0 degrees above the horizontal from the top edge of a cliff with an initial speed of 15 m/s. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom at 6.30 s. What is the height of the cliff? (Assume no air resistance and that a = g = 9.81 m/s.)

Height of cliff = 147.43 m

Explanation:

We have equation of motion s = ut + 0.5 at²

Consider the vertical motion of stone and up as positive,

Initial velocity, u = 15sin30=7.5m/s

Time taken, t = 6.30 s

Acceleration, a = - 9.81 m/s²

Substituting

s = ut + 0.5 at²

s = 7.5 x 6.30 + 0.5 x (-9.81) x 6.30²

s = - 147.43 m

So the stone goes below 147.43 m below the initial postilion

Height of cliff = 147.43 m