(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A rock is shot vertically upward form the edge of the top of a tall building. The rock reaches its maximum heigh above the top of the building 1.08s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 6 s after it was launched.

With what upward velocity was the rock shot? 10.58 m/s

What maximum height above the top of the building is reached by the rock? ____ m

How tall is the building? _______ m

2. Relevant equations

x(t)= initial position + final velocity * time

v(t)= (acceleration * time) + initial velocity

x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

x= initial position * (average velocity * time)

average velocity= (final velocity - initial velocity) / (2)

(final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position

3. The attempt at a solution

I can't figure out how to work this problem. I got 58.8 m/s as my final velocity on the second half. I tried solving for change in x by squaring final velocity and dividing by 19.6 (becuase 2 * 9.8 m/s^2). Can anyone explain how these two problems are worked?

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# Homework Help: Velocity, height, etc

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