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

Part A:

If the faster stone takes 8.00 s to return to the ground, how long will it take the slower

stone to return? t = ?

Part B:

If the slower stone reaches a maximum height of H, how high (in terms of H) will the faster stone go?

A) Vf = Vo - g*t

Vf = 0

g = 9.8 m/s^2

Vo = g * t = > t up = Vo / g

t total = 2*t = 2*Vo / g

t slow = 2*Vo / g

t fast = 2*[3Vo] / g = 6Vo / g = > Vo = g * t fast / 6 = 9.8 m/s^2 * 8.00 s / 6 = 13.07 m/s = Vo

t slow = 2 * 13.07 m/s / 9.8 m/s^2 = 2.67 s.

B)

Vf ^2 = Vo ^2 - 2gy

Vf = 0 = > Vo ^2 = 2gy

Vo ^2 = 2gH

[3Vo]^2 = 2gy

=> [3Vo]^2 / Vo^2 = 2gy / 2gH

=> 9 = y/H = > y = 9H