Posted by **Anonymous** on Friday, November 21, 2008 at 1:20pm.

A boy stands on the edge of a building 7 meters high and throws a rock at 60-degree angle. The release point from his hand is 1.6 meters above his feet(so the release point above the ground is 8.6 meters.) The rock lands on the ground 15 meters away from the building at a point level with the base of the building. Assuming no air resistance, what was the initial velocity of the throw? (assume gravity is 9.81m/s^2)

I'm stuck on how to solve. I think I need to figure out how long the rock is in the air, which would be time it takes for the rock to arrive at the apex of the throw and then drop from the apex height to the ground. But since I don't know the initial velocity, how do I determine the height of the apex?

- Physics -
**drwls**, Friday, November 21, 2008 at 1:51pm
The horizontal distance away that it landed after T seconds tells you that

Vo cos 60 T = 15 , which leads to

T = 30/Vo

Also, the vertical velocity component Vo sin 60 is related to the total time T in the air by

Vo sin 60 T - (g/2) T^2 = -8.6

You have two equations in two unknowns.

Substitue 30/Vo for T in the last equation and solve for Vo

- Physics -
**Anonymous**, Friday, November 21, 2008 at 2:55pm
thanks, I get it.

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