# Motion in a straight line HELP

• goaliejoe35
In summary, the problem is asking for the height of a rock that is dropped from a building 1.2 seconds before it reaches the ground. Using the formula y=(1/2)gt^2, we can find the time it takes for the rock to reach the ground, which is 4.23 seconds. Then, by subtracting 1.2 seconds from the total time, we can find the time the rock falls before reaching the desired height. Plugging this time into the formula y=(1/2)gt^2, we get a displacement of 7.056 meters. However, this is the height from the top of the building. To find the height from the ground, we subtract this displacement from

#### goaliejoe35

Motion in a straight line HELP!

Here's my problem...

A rock is dropped (from rest) from the top of a 87.5-m-tall building. How far above the ground is the rock 1.2 s before it reaches the ground?

...so far this is what I came up with

y=(1/2)gt^2

t=sqrt(2y/g) = sqrt((2*87.5)/9.8) = 4.23 s <-- I believe this is the time for the entire drop?

Then I subtracted 4.23-1.2 = 3.03 s <---- so that's the time the rock falls before the point at which i need to calculate the height.

After that i get lost and can't seem to come up with a sensible answer.

If someone could walk me through this I'd really appreciate it!

well you are going correct with the formula

$$s=ut+\frac{1}{2}at^2$$

since it starts at rest,u=0. and the acceleration a=g

$$s=\frac{1}{2}gt^2$$

so that is correct.

you want to find s,when t=1.2. Just put it into the formula and you'll get s when t=1.2

OK so if I plug 1.2 s into the equation I get this...

S=(1/2)*9.8(1.2)^2
= 7.056 m/s

Correct?

Now I still don't get how I figure out how high off the ground the rock is at 1.2 seconds?

goaliejoe35 said:
OK so if I plug 1.2 s into the equation I get this...

S=(1/2)*9.8(1.2)^2
= 7.056 m/s

Correct?

Now I still don't get how I figure out how high off the ground the rock is at 1.2 seconds?

s is displacement. so the unit is m.

So 7.056 is the height from the top of the building to the point when t=1.2seconds.

The height of the entire building is 87.5m.

So the height from the ground would just be (height of building)-(distance when t=1.2)

Hey goaliejoe,
Your on the right track but use your time that you found (3.03) and plug that into the formula
Y = Yo - (1/2)at^2. Your y naught is the starting height.
:)