Calculating the Time and Height of Falling Stones: A Kinematics Problem

Just before they hit the ground" means the time just before they hit. Since you know the height you can calculate the velocity just before they hit. (Actually, just after they are dropped. As soon as they are released they start to fall and their velocity increases until they hit the ground. "Just before" the time they hit the ground is the instant they are released.) That is, if you take the time calculated, 3.88 seconds, you can calculate the velocity 3.88 seconds after the stones were released. Since they are at the top of their flight at that time, that will be the maximum velocity. Just use the original equation with t= 3.88
  • #1
Matt
Ok, so there are two stones. One is dropped (not thrown) off a building. 2 seconds later a rock is thrown at 30 m/s. The rocks hit the ground at the same time. Here's what I have so far:

R1:

v0 = 0 m/s
T = unknown
x = o m
x0 = unknown
a = 9.8 m/s^2

R2:

v0 = 30 m/s
T = Time of R1 - 2 s
x = o m
x0 = unknown
a = 30 m/s @ 9.80 m/s^2

I know that these two equations must equal each other:

(.5)(9.80 m/s^2)(T)^s = (30m/s)(T-2) + (.5)(9.80m/s^2)(T-2)^2

However, I have expanded this a million times and for some reason I can't get things to click. I mean, I could do guess and check but there has got to be an easier way...

The question asks how long it took the first stone to reach the gound, how high the building is, and what speeds the stones are at before they hit the ground.

Since they ask for the time first, I thought I would have to use the 2nd kinematic equation. But then I realized that I don't know what the initial height was. I feel stuck because I have these two variables the initial height and time. What do I use to find these one at a time? Lol, I'm stuck.
 
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  • #2
Originally posted by Matt
(.5)(9.80 m/s^2)(T)^s = (30m/s)(T-2) + (.5)(9.80m/s^2)(T-2)^2

However, I have expanded this a million times and for some reason I can't get things to click. I mean, I could do guess and check but there has got to be an easier way...

There is an easy way: The quadratic formula.

Expand out the square, and you'll see that you have a simple quadratic polynomial. You can then solve by:

x=(-b±(b2-4ac)1/2)/2a
 
  • #3
Comes down to simplfied before plugging into equation:

(-19.6 m/s^2)T + (30 m/s)T + (19.6 m/s^2) - (60 m/s)

How does this follow std form?
 
  • #4
You messed up somewhere, because you should have a "T2" term in there. Slow down and have another crack at it.
 
  • #5
(-19.6 m/s^2)T + (30 m/s)T + (19.6 m/s^2) - (60 m/s)

You clearly mean T2 in the first term because you will need seconds squared to cancel the s^2 term in -19.6 m/s^2.

For the same reason "(19.6 m/s^2)- (60 m/s)" is impossible- you can't add (or subtract) velocity and acceleration.

Finally, to solve an equation, you have to have an equation! What is this supposed to be equal to?
 
  • #6
Ok, I was looking at my physics book a little more...

You can find time with the quadratic equation with this equation...

y = y0 + v0t + .5at^2

However, you cannot solve for time with this method unless you have all the information except time. In my case, this is what I have for the second stone:

x = 0
x0 = unknown
t = unknown
a = 9.80 m/s
v0 = 30 m/s (it was thrown not dropped)

Due the fact that there are two unknowns I cannot solve with the quadratic. The other stone, which was dropped, must be included in the mix somewhere - it has to be a two part problem that I can't see.

First stone:

x = 0
x0 = unknown
t = unknown
a = 9.0 m/s^2
v0 = 0 m/s

If you think about it these two equations must equal each other if the two rocks hit the ground at the same time:

(.5)(9.80 m/s^2)(T)^2 = (30m/s)(T-2) + (.5)(9.80m/s^2)(T-2)^2

I have the actual answer, its in the back of the book, and if you plug the time into it, the equations equal each other. However, I have expanded this equation a million times. the (4.90 m/s^2)T^2 always cancels out (which doesn't give rise to std form for the quadratic equation. I'm really stuck guys and gals.
 
  • #7
Okay, your equation is:

4.9T2= 4.9(T-2)2+ 30(T-2)

multipling that out:

4.9T2= 4.9(T2- 4T+ 4)+ 30T- 60
= 4.9T2- 19.6T+ 19.6+ 30T- 60

The "4.9T2" terms on both sides cancel! We don't have a quadratic equation after all!

We do have, after we move the terms involving T to one side,
we have 30T- 19.6T= 60- 19.6 or 10.4T= 40.4 so T= 40.4/10.4= 3.88 seconds. Now, you can use that value of T to find the height of the building.
 
  • #8
Is it ok that you neglected units like that? Like m/s^s and m/s and s?
 
  • #9
Well, I guess you can, because according to the book that answer is correct. It was easy to find the height based off of this, but what does "what speeds the stones are at before they hit the ground" mean? "Just before" could be ay any point couldn't it?
 
  • #10
Oh wait I understand it now, when they say right before they mean the maximum velocity of the first and second stones. That is a stupid way for the book to word it, lol.
 
  • #11
First, I didn't need to carry the units along because I had checked them and knew they were correct.

Second, any time a textbook seems to word something in a "stupid way" look again- you might learn something. (I'm not saying it CAN'T be stupid- just that authors (and editors) tend to be very careful about wording.)

In this case, if they had said "maximum velocity" that would have been assuming things not yet given. The point of "just before" (I notice you say "before they hit the ground" and "just before"- which was it?) is that you can use the time you have already calculated for hitting the ground. AT the time they hit the ground they start decellerating rapidly so that would be ambiguous.
 

What is the formula for calculating the time and height of falling stones?

The formula for calculating the time and height of falling stones is given by h = 1/2gt^2, where h is the height, g is the acceleration due to gravity, and t is the time.

How do you determine the acceleration due to gravity?

The acceleration due to gravity can be determined using the formula g = F/m, where F is the force of gravity and m is the mass of the object. On Earth, the value of g is approximately 9.8 m/s^2.

What are the units for time and height in this problem?

The units for time and height in this problem depend on the unit system being used. In the metric system, time is measured in seconds and height is measured in meters. In the imperial system, time is measured in seconds and height is measured in feet.

Can this formula be used for any object falling from any height?

Yes, this formula can be used for any object falling from any height as long as the object is in free fall and there are no external forces acting on it (such as air resistance).

How accurate is this formula in real-life situations?

This formula provides a simplified model for calculating the time and height of falling objects. In real-life situations, factors such as air resistance and the shape of the object may affect the accuracy of the results. However, for most practical purposes, this formula provides a good estimate of the time and height of falling objects.

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