Question

You stand on top of a cliff facing a perfectly flat valley below. In one hand is a stone and in the other a stopwatch. You throw the stone at 30.0 m/s and an angle of 30.0° above the horizontal. You time the stone with your stopwatch and find that it lands 6.91 seconds after you threw it.

A)How high above the valley floor was the stone when you threw it?
y0 = ____m

B)At what horizontal distance (x) from the cliff did the stone land?
x= ___ m

C)What was the magnitude of the stone's velocity just before it hit the ground?
v=___ m/s

D)Graph the the x-velocity vx versus time for the motion of the stone. Then choose the correct graph below.

E)Graph the the y-velocity vy versus time for the motion of the stone. Then choose the correct graph below.

Answer 1

Height of cliff = 147.43 m

Step-by-step 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