Part A A small rock is thrown vertically upward with a speed of 18.0 m/s from th

Question

Part A A small rock is thrown vertically upward with a speed of 18.0 m/s from the edge of the roof of a 37.0 m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. What is the speed of the rock just before it hits the street? Express your answer with the appropriate units. yValue Units Submit Request Answer Part B How much time elapses from when the rock is thrown until it hits the street? Express your answer with the appropriate units. tValue Units

Explanation / Answer

Givens:
initial velocity, Vi = 1280m/s
acceleration (due to gravity), a = -9.81m/s^2
initial height, hi = 37.0m
final height, hf = 0m
final velocity, Vf = ?
time, t = ?

To solve for Vf, I would choose the kinetic equation:
Vf^2 - Vi^2 = 2a(hf - hi)
Vf^2 - (18.0m/s)^2 = 2(-9.81m/s^2)(0m - 37.0m)
Vf^2 - 324m^2/s^2 = 725.2m^2/s^2
Vf^2 = 1049.2m^2/s^2
Vf = SQRT(1049.2m^2/s^2)
Vf = +/- 32.391m/s

If we were talking about velocity (a vector), we would reject the positive value and choose the negative one, because we know the rock would be heading downwards. The question, however, asked for a speed, and speed is not a vector (direction doesn't matter), so the speed is (positive) 28.5m/s

To solve for t, I would then choose
Vf = Vi + at
=> Vf - Vi = at
=> (Vf - Vi)/a = t
t = (-32.391m/s - 18m/s)/(-9.81m/s^2)
t = (-50.291m/s)/(-9.81m/s^2)
t = 5.126s

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