Motion in a straight line HELP

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
  • #1
goaliejoe35
72
0
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!
 
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  • #2
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
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)
 
  • #5
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.
:)
 

1. What is motion in a straight line?

Motion in a straight line refers to the movement of an object along a single, straight path. This type of motion is often described in terms of distance, displacement, speed, and velocity.

2. What is the difference between distance and displacement?

Distance is the total length of the path an object has traveled, while displacement is the straight-line distance between an object's starting and ending point.

3. How is speed different from velocity?

Speed is a measure of how fast an object is moving, while velocity is the speed of an object in a specific direction. Therefore, velocity takes into account the object's speed and direction.

4. How is acceleration related to motion in a straight line?

Acceleration is the rate of change of an object's velocity. In motion in a straight line, acceleration can be positive (speeding up) or negative (slowing down).

5. What are the different types of motion in a straight line?

The three main types of motion in a straight line are uniform motion, uniformly accelerated motion, and non-uniform motion. Uniform motion has a constant speed, while uniformly accelerated motion has a constant acceleration. Non-uniform motion has a changing speed or acceleration.

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