Two stones are thrown vertically upward from the ground, one with 3 times the initial speed of the other. Assume free fall.

a) If the faster stone takes 12.0s to return to the ground, how long will take the slower stone to return?

b) if the slower stone reach a max height of H, how high (in term of H) will the faster stone go?

We are given the data two stones thrown at different velocities results to different times. We can use the equation y = vot - 1/2 gt2. At maximum height, dy/dt = 0 = vo - gt. If t2 is 12 s, then vo is equal to 117.6 m/s. Then, the time of t1 is equal to 117.6*3/9.8 equal to 36 seconds. When t = 36 seconds, then y = 117.6*36 - 1/2 * 9.8*36^2 equal to - 2116.8 m