Calculating Distance of a Thrown Rock from a Building | Physics Problem Solution

[yadda]

hello. my physics teacher gave us this problem.
someone is standing on a building, and throws a rock. the height from which the rock is thrown is 230 m. the initial velocity of the rock is 100 m/sec. thrown in the positive x direction, how far will the rock go by the time it hits the ground?

now, I used some equations to try to figure this out. I got an answer - 508.684 m. But, I'm pretty sure this must be a wrong answer. (so far, I usually get most of my homework wrong. I always miss something) it seemed too easy. and I didn't use the height from which the rock was thrown in any way. should I have needed to? I know I'm missing something here. please help me out. thanks.

 

[chroot]

Why don't you first tell us how you approached the problem? What reasoning did you use to do the things you did?

- Warren

 

[BoulderHead]

hint:

Can you figure how long it would take the rock to hit the ground if it were dropped straight down from that distance?

 

[yadda]

Warren ~ at class, we worked out a problem concerning how high the rock would go if thrown straight up. the equations we used were these: V(t) = Vo + at ...(V:velocity, t:time, Vo:velocity at zero distance, and a:acceleration)
X = Xo + Vo(t) + 1/2 a(t)squared ...(X:distance, and 1/2 a(t)squared because of constant acceleration)

Our teacher didn't give us any other equations to use, so I thought what the heck and went ahead and tried them. here's what I did:
V(t) = Vo + at
0 = 100m/s + (-9.81m/sec2)t ...(negative because of deceleration)...
9.81t = 100 (divide both by 9.81)
t = 100/9.81 = 10.1937 sec

X = Xo + Vo(t) + 1/2 a(t)squared
X = 0 + 100t + 1/2 (-9.81)(10.1937)squared
X = 100(10.1937) + (0.5)(-9.81)(103.9115)
X = 1,019.37 - 509.6859
X = 509.684m -that's the final answer. but tell me - I shouldn't have done it that way, right?

BoulderHead ~ I can figure that out, but can you tell me how that would help me?

 

[PrudensOptimus]

I think you had a misconception of V(t) = 0.


When V(t) = 0, it is implying when the object reaches it's maximum point.


In this case, I think you should use Δy = -0.5gt^2 and find at the point Δy = -230 m.

You should get t = sqrt(46) or about 6.8s.

x = v(initial) * t.

= 100 * 6.8 = 680m.

 

[Doc Al]

Looks like you're plugging away at random, mixing horizontal and vertical components of motion. You are using equations only valid for accelerated motion, which only apply to the vertical component.

Consider the horizontal and vertical directions separately.
What's the initial speed in the horizontal direction?
What's the initial speed in the vertical direction?

So... think about BoulderHead's hint. Does an initial horizontal speed affect the time it takes for an object to fall?

 

[BoulderHead]

Yadda,
What is going on with a problem like this is that in order to know how far the rock travels (horizontal, x-axis movement) you need to know how fast it is moving (which was provided) as well as the length of time it will be traveling. So, unless you are able to determine how long the rock is in the air you won't be able to determine how far it travels, because the element of time is unknown. How long will it be in the air? …Until gravity pulls it down to the earth.

 

[yadda]

yeah, Doc ~ you hit the button on the nose. In stuff like this, I'm kinda at a loss... I get it if I've been told exactly what to do, but if the hint is too subtle, with something like this, I don't get it.

but I think I might be starting to understand... (thanks!) mmm. but about BoulderHead ~ that hint was about free falling. if I figure out the free fall time, it won't be the same time as if the rock was thrown with a 100m/s force, right? mmm... but could I use that formula? .. V=9.76t, so.. er..no, right? I have a feeling I'm headed in the wrong direction again.
(sorry if this wears on your patience, but I'm truly distressed about this. I'm just having trouble understanding. really, thank you for your help so far.)

once I figure out the time, can I still use the x formula that I have - to then figure out the distance? is that a right equation for this?

 

[BoulderHead]

Originally posted by yadda
BoulderHead ~ that hint was about free falling. if I figure out the free fall time, it won't be the same time as if the rock was thrown with a 100m/s force, right?

Ahhh, you're fixing to stumble onto something you may find interesting; for it depends on the angle the rock was thrown. If the rock were thrown straight up it would take longer for it hit the Earth than if it were thrown straight down, correct? But in both cases the force due to gravity is constant. The force due to gravity isn’t going to change just because the rock is thrown parallel with the surface of the Earth either, and isn’t this exactly what is happening?

Re-read what Doc Al said in middle and bottom parts of his reply.

 

[yadda]

ok. the initial speed in the horizontal direction is 100m/s. the initial speed in the vertical direction is 0. (I'm getting this right?)mmm... so if we have an x line, the only thing that will affect whether an item's speed is longer or can be calculated the same as a free falling object is if it goes above the line, or below the line? going in the direction of the line doesn't affect it at all.. it will only be affected if it's going above the line. but since we know the object starts at a certain point and only goes down from there, I can use the free falling equation? such as: V=9.76t ? my V is 100m/s, so that would be 100=9.76t, so t= 10.2459sec . Will this get me to the correct answer?

I was looking at what everyone said. now, PrudensOptimus gave me an equation. it doesn't seem to coincide with what you're hinting at... if I'm getting it right. Can I use the above equation? if yes, how can PrudensOptimus equation be correct? if no, then I'm still at a loss. I copied this from another thread in this forum:
____________________

A stone is thrown horizontally at a speed of 10 m/s from the top of the cliff 78.4 cm high. How long does it take the stone to reach the bottom of the cliff? How far from the base of the cliff does the stone strike the ground

I assume you meant to say that you know that "vertical" and horizontal motion are independant.

The amount of time the stone remains in the air depends entirely on the vertical motion. It would take the same amount of time to hit the ground as if it were dropped. SO, can you calculate how long it would take a stone to drop straight down? THis the amount of time your stone is in the air, and so it is also the amount of time it is moving at 10 m/s in the horizontal direction. Now you have v and t. FInd d.
______________________________________________ - is this what you mean, BoulderHead?... (is that even what I did?)

about PrudensOptimus - I don't understand, how do you get from Äy = -0.5gt^2 to t=sqrt(46) ?

 

[PrudensOptimus]

Originally posted by yadda
ok. the initial speed in the horizontal direction is 100m/s. the initial speed in the vertical direction is 0. (I'm getting this right?)mmm... so if we have an x line, the only thing that will affect whether an item's speed is longer or can be calculated the same as a free falling object is if it goes above the line, or below the line? going in the direction of the line doesn't affect it at all.. it will only be affected if it's going above the line. but since we know the object starts at a certain point and only goes down from there, I can use the free falling equation? such as: V=9.76t ? my V is 100m/s, so that would be 100=9.76t, so t= 10.2459sec . Will this get me to the correct answer?

I was looking at what everyone said. now, PrudensOptimus gave me an equation. it doesn't seem to coincide with what you're hinting at... if I'm getting it right. Can I use the above equation? if yes, how can PrudensOptimus equation be correct? if no, then I'm still at a loss. I copied this from another thread in this forum:
____________________


I assume you meant to say that you know that "vertical" and horizontal motion are independant.

The amount of time the stone remains in the air depends entirely on the vertical motion. It would take the same amount of time to hit the ground as if it were dropped. SO, can you calculate how long it would take a stone to drop straight down? THis the amount of time your stone is in the air, and so it is also the amount of time it is moving at 10 m/s in the horizontal direction. Now you have v and t. FInd d.
______________________________________________ - is this what you mean, BoulderHead?... (is that even what I did?)

about PrudensOptimus - I don't understand, how do you get from Äy = -0.5gt^2 to t=sqrt(46) ?

#1. You used v = 9.8t wrong. You mentioned initial velocity to be 100m/s. The v in v = 9.8t is the FINAL VELOCITY, not initial. The full equation should be v = v₀ + at, where a = -9.8m/s^2 for free falling objects.

#2. In the problem, it gave you Δy = -230 meters. So by using Δy = -0.5gt^2, and find t.

 

[chroot]

Hi yadda,

This has gone on long enough. Let's break the problem down into pieces. See if you can follow the pieces, and do each one in turn. You'll get the final answer if you do each of the pieces correctly.

1) Figure out how long the rock takes to hit the ground. This is only dependent upon its initial height off the ground and its initial vertical velocity, which is zero. It has nothing to do with its horizontal motion.

The equation you need is:

s(t) = v_0y t + 1/2 a t²

230 = 0 + 0.5 * 9.8 * t²

Solve for t.

2) The rock is moving 100 m/s horizontally during the entire duration of its flight. Remember that its horizontal and vertical motions are independent, and gravity only affects the vertical motion.

The horizontal distance can be found with the equation:

s(t) = v_0x t

s(t) = 100 t

Plug in the t you found in part 1, and solve for s(t).

- Warren

edit: thanks Prudens

 

[PrudensOptimus]

Originally posted by chroot

230 = 0 - 0.5 * 9.8 * t²

- Warren [/B]

he forgot that 230 is -230.

 

[yadda]

yeah, I noticed that...