Calculating Velocity and Maximum Height of a Rock Launched from a Tall Building

[lauriecherie]

Homework Statement

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

Homework 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

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?

 

[Redbelly98]

A hint for the 2nd question:
You know the acceleration and time, and also the initial velocity (from doing the 1st question). Which equation(s) can make use of all those bits of information, and give you the height reached by the rock?

 

[lauriecherie]

I've been working on it again. Is 58.8 m correct for the maximum height?

 

[Delphi51]

Your initial velocity looks good.
Why can't you use your x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position
to find the maximum height above the starting point?

It should work for the building height, too - just call the initial height zero and you'll get a nice negative answer for x at time 6. No need to find the final velocity.

 

[LowlyPion]

The height can be worked out simply by observing that X = 1/2*g*t²

You have the time, ... there's your height. (The reason you can do this is because an object uniformly accelerating from 0 to some speed it the same as it decelerating over that same time in going up. Conservative fields and all that.)

Using the same relationship but with 6 seconds tells you the height from Max height. That distance less the first distance ...

 

[lauriecherie]

I keep coming out with 176.4 m, which Webassign says is incorrect. Argghhh... Am I doing something else wrong?

 

[LowlyPion]

I keep coming out with 176.4 m, which Webassign says is incorrect. Argghhh... Am I doing something else wrong?

You haven't subtracted the distance it went up first.

 

[lauriecherie]

You haven't subtracted the distance it went up first.

Ok so I now subtracted 5.71 from 174.60 and got 168.89 m as the height of the building. It is still marking it incorrect. I'm so confused.

 

[LowlyPion]

You also need to subtract the 1.08 from 6s because it was 6s after launch.

 

[lauriecherie]

SOLVED! Thanks for all the help!