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Motion in a straight line HELP

  • #1
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 thats 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!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
well you are going correct with the formula

[tex]s=ut+\frac{1}{2}at^2[/tex]


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

[tex]s=\frac{1}{2}gt^2[/tex]


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
 
  • #3
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?
 
  • #4
rock.freak667
Homework Helper
6,230
31
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)
 
  • #5
4
0
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.
:)
 

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