physics help plz?

Question

A stone is thrown at an angle of 30.0° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone’s trajectory time from the top of the cliff to the bottom at 5.60 s. What is the height of the cliff? (Assume no air resistance and that a ^y= -g = -9.81 m/s^2.)

Answer 1

Start by finding the initial speed in the horizontal (x) direction and the vertical (y) direction:

vx = v0 *cos(q) = 12m/s*cos(30) = 6*sqrt(3) m/s

vy = v0*sin(q) = 6 m/s

Now the displacment in y is given by

y = 1/2*a*t^2+ vy*t +y0 where a = acceleration, t = time and y0 = initial height.

Let the downward direction have a negative sign, then a = -g = -9.8 m/s^2 and vy = +6m/s.

Let all vertical displacement be measured from the bottom of the cliff so that y = 0. Then:

0 = -1/2*(9.8m/s^2)*t^2 + 6 (m/s)*t +y0 --> y0 = height of cliff

y0 = 1/2*(9.8 m/s^2)*t^2 - 6m/s*t

You are given t = 5.6 s. Subsituting into above equation:

y0 = 120 m