SOLUTION: A small rock is thrown vertically upward with a speed of 18 m/s from the edge of the roof of a 30.0 m-tall building. The rock doesn't hit the building on its way down and lands in
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Question 1040878: A small rock is thrown vertically upward with a speed of 18 m/s from the edge of the roof of a 30.0 m-tall building. The rock doesn't hit the building on its way down and lands in the street below. Air resistance can be neglected. (a.) What is the speed of the rock just before it hits the street? (b.) How much time elapses from when the rock is thrown until it hits the street? Found 2 solutions by addingup, ikleyn:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! At different points on Earth, objects fall with an acceleration between 9.78 and 9.83 m/s2 depending on altitude and latitude, with a conventional standard value of exactly 9.80665 m/s2 (approximately 32.174 ft/s2).
Let's round off the conventional standard and call it 9.81 m/s^2
:
v^2 = u^2+2as
v^2 = 18^2+2*9.81*30 = 912.6
v = sqrt912.6 = 30.21
:
v = u+at
30.21 = -18+9.81t
9.81t = 48.21t
t = 4.914 (if you take it to 3 decimal places)
:
The velocity (v) just before the rock hits the ground is 30.21m/s
The tiime that the rock is in flight is 4.914 seconds
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The formulas and the process are correct. Check the calculations, I went fast.
John
