Dynamics: Projectile Motion velocity equation

In summary, the scenario as stated does not specify at what point (height) the projectile is being released with respect to the tunnel's ceiling (or floor) at point (A). This lack of information affects the calculation of the projectile's rise time and required launch velocity (u) to reach point (B).
  • #1
cojaro
9
0
1. "A projectile is fired with a velocity u at the entrance A of a horizontal tunnel of length L and height H. Determine the minimum value of u and the corresponding angle [itex] \theta [/itex] for which the projectile will reach B at the other end of the tunnel without touching the top of the tunnel."

[itex]
\begin{center}
v_0 = u\\
x = L , x_0 = 0\\
y = H , y_0 = 0\\
\end{center}
[/itex]
The book gives the answer as http://latex.sidoh.org/?render=%24u%20%3D%20%5Csqrt%7B2gH%7D%20%5Csqrt%7B1%20%2B%20%5Cleft%28%20%5Cdfrac%7BL%7D%7B4H%7D%20%5Cright%29%5E2%24 and http://latex.sidoh.org/?render=%24%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%7B%5Cdfrac%7B4H%7D%7BL%7D%7D%5Cright%29%24

Homework Equations


http://latex.sidoh.org/?render=%5Cbegin%7Bcenter%7D%0D%0A%24v_%7Bx_0%7D%3Du%5Ccos%7B%5Ctheta%7D%3Dv_x%24%20%5C%5C%0D%0A%24v_%7By_0%7D%3Du%5Csin%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%24x%3Dut%5Ccos%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%24y%3D-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%2But%5Csin%7B%5Ctheta%7D%2By_0%24%20%5C%5C%0D%0A%24v_y%3D-gt%2Bu%5Csin%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%5Cend%7Bcenter%7D

The Attempt at a Solution



So, I asked my teacher to set it up for me. I've wrapped my head around this for the past week and just don't see it! I've tried so many danged angles of approach and still get nowhere. My teacher's setup didn't help me much, either, so there's not much work-out to do. My teacher left me with four equations:

http://latex.sidoh.org/?render=%5Cbegin%7Bcenter%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0Ax%20%3D%20%5Cdfrac%7BL%7D%7B2%7D%20%3D%20u%20%5Ccos%7B%5Ctheta%7D%28%20%5Cfrac%7Bt_f%7D%7B2%7D%29%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0AL%20%3D%20u%20t_f%20%5Ccos%7B%5Ctheta%7D%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0AH%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20g%20%5Cleft%28%5Cfrac%7Bt_f%7D%7B2%7D%5Cright%29%5E2%20%2B%20u%20%5Csin%7B%5Ctheta%7D%28%20%5Cfrac%7Bt_f%7D%7B2%7D%29%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0A0%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20g%20t_f%5E2%20%2B%20u%20t_f%5Csin%7B%5Ctheta%7D%0D%0A%5Cend%7Bequation%7D%0D%0A%0D%0A%5Cend%7Bcenter%7D

with [itex]t_f[/itex] being the time to get to B. He told me to plug'n'play and that I'd eventually get the answer.

(I also had some troubles writing latex using the [ latex] XXX [ \latex ] HTML things. It kept messing up, misaligned and showing the variables as text, not math text, namely, v versus [itex] v [/itex]. )
 
Last edited by a moderator:
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  • #2
with tf being the time at maximum H, assuming the path is perfectly symmetric.

I'm not sure I agree with you there--tf should actually be the time when the projectile has traveled the whole lenght, L.

This is the sort of problem you'll have to play around with for a while and I can't say I've got a solution for you, but if I were to do the problem I would try to start with finding an expression for tf, since you don't want it left in the answer.
 
  • #3
Hannisch said:
I'm not sure I agree with you there--tf should actually be the time when the projectile has traveled the whole lenght, L.

Yeah, you're right. I missed that. My mind is blahhhh after this week.
 
  • #4
So in my continuous efforts, I realized that [tex]u = \sqrt{v_y^2 + v_x^2}[/tex] so it's clear that [tex]v_y^2 = 2gH[/tex] and, with a little work, that [tex]v_x^2 = \dfrac{gL^2}{8H}[/tex], but that's as far as I can get. Working backwards really doesn't help all that much. I've been trying to find where the L/4H comes from, but so far all efforts have been to no avail =[
 
  • #5
L/4H looks kind of like it's 1/tan(x)

Since tan(x)=4H/L.

I've been trying to do this and all I've come to that tan(x)=8H/(4L-L2) and the conclusion that perhaps a Saturday evening isn't the time when my mind is the most set up to do physics.
 
  • #6
cojaro said:
1. "A projectile is fired with a velocity u at the entrance A of a horizontal tunnel of length L and height H. Determine the minimum value of u and the corresponding angle [itex] \theta [/itex] for which the projectile will reach B at the other end of the tunnel without touching the top of the tunnel."

[itex]
\begin{center}
v_0 = u\\
x = L , x_0 = 0\\
y = H , y_0 = 0\\
\end{center}
[/itex]
The book gives the answer as http://latex.sidoh.org/?render=%24u%20%3D%20%5Csqrt%7B2gH%7D%20%5Csqrt%7B1%20%2B%20%5Cleft%28%20%5Cdfrac%7BL%7D%7B4H%7D%20%5Cright%29%5E2%24 and http://latex.sidoh.org/?render=%24%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%7B%5Cdfrac%7B4H%7D%7BL%7D%7D%5Cright%29%24

Homework Equations


http://latex.sidoh.org/?render=%5Cbegin%7Bcenter%7D%0D%0A%24v_%7Bx_0%7D%3Du%5Ccos%7B%5Ctheta%7D%3Dv_x%24%20%5C%5C%0D%0A%24v_%7By_0%7D%3Du%5Csin%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%24x%3Dut%5Ccos%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%24y%3D-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%2But%5Csin%7B%5Ctheta%7D%2By_0%24%20%5C%5C%0D%0A%24v_y%3D-gt%2Bu%5Csin%7B%5Ctheta%7D%24%20%5C%5C%0D%0A%5Cend%7Bcenter%7D

The Attempt at a Solution



So, I asked my teacher to set it up for me. I've wrapped my head around this for the past week and just don't see it! I've tried so many danged angles of approach and still get nowhere. My teacher's setup didn't help me much, either, so there's not much work-out to do. My teacher left me with four equations:

http://latex.sidoh.org/?render=%5Cbegin%7Bcenter%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0Ax%20%3D%20%5Cdfrac%7BL%7D%7B2%7D%20%3D%20u%20%5Ccos%7B%5Ctheta%7D%28%20%5Cfrac%7Bt_f%7D%7B2%7D%29%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0AL%20%3D%20u%20t_f%20%5Ccos%7B%5Ctheta%7D%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0AH%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20g%20%5Cleft%28%5Cfrac%7Bt_f%7D%7B2%7D%5Cright%29%5E2%20%2B%20u%20%5Csin%7B%5Ctheta%7D%28%20%5Cfrac%7Bt_f%7D%7B2%7D%29%0D%0A%5Cend%7Bequation%7D%0D%0A%5Cbegin%7Bequation%2A%7D%0D%0A0%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20g%20t_f%5E2%20%2B%20u%20t_f%5Csin%7B%5Ctheta%7D%0D%0A%5Cend%7Bequation%7D%0D%0A%0D%0A%5Cend%7Bcenter%7D

with [itex]t_f[/itex] being the time to get to B. He told me to plug'n'play and that I'd eventually get the answer.

(I also had some troubles writing latex using the [ latex] XXX [ \latex ] HTML things. It kept messing up, misaligned and showing the variables as text, not math text, namely, v versus [itex] v [/itex]. )

There’s actually a problem with the scenario as stated.

We understand that (u) will represent the velocity of the projectile and (A) represents the entrance of a horizontal tunnel whose dimensions are given by Length (L) and Height (H) with (B) representing the end of the tunnel.

Here’s the problem

The height at which the projectile is being released has not been given. For instance, the projectile might be released from ground level thereby giving you maximum rise time, but it could just as easily be released from any height in between the tunnel’s floor and ceiling at point (A), which ultimately alters the amount of rise time of the projectile and the required launch velocity (u). The less rise time given to the projectile, the greater will be the required launch velocity (u) to make the projectile reach point (B).

Since it makes a difference, I’d have your teacher clarify at what point (height) the projectile is being released with respect to the tunnel’s ceiling (or floor) at point (A). After all, very few projectiles are released from a point that is level with the ground. :wink:
 
Last edited by a moderator:
  • #7
Gnosis said:
There’s actually a problem with the scenario as stated.

We understand that (u) will represent the velocity of the projectile and (A) represents the entrance of a horizontal tunnel whose dimensions are given by Length (L) and Height (H) with (B) representing the end of the tunnel.

Since it makes a difference, I’d have your teacher clarify at what point (height) the projectile is being released with respect to the tunnel’s ceiling (or floor) at point (A). After all, very few projectiles are released from a point that is level with the ground. :wink:

I guess I wasn't clear. =\ I hate this problem. The object's fired from [tex]H=0[/tex] towards [tex]H=H_{max}[/tex] at an angle [tex]\theta[/tex].
 
  • #8
It does say y0=0, so you were quite clear, actually. But that doesn't bring you further towards the answer, right?
 
  • #9
Hannisch said:
It does say y0=0, so you were quite clear, actually. But that doesn't bring you further towards the answer, right?

No, not at all =[ I've filled quite a few pages of notebook paper just trying different approaches and solutions, working backwards from the answer, etc. I'm not making it more complicated that it really is, am I? I tend to do that.
 
  • #10
Hannisch said:
It does say y0=0, so you were quite clear, actually. But that doesn't bring you further towards the answer, right?

'Yo' wasn't defined, so it could represent anything, and since 'Xo' was also included and undefined, they could represent initial Y and X velocity components, which are entirely different than the point (height) of release of the projectile. Only a fool assumes that which is not given as a certainty.
 
  • #11
Gnosis said:
'Yo' wasn't defined, so it could represent anything, and since 'Xo' was also included and undefined, they could represent initial Y and X velocity components, which are entirely different than the point (height) of release of the projectile. Only a fool assumes that which is not given as a certainty.

[tex]y_0[/tex] and [tex]x_0[/tex] were defined. And I'd never let y or x represent velocity components. I use [tex]v_{y_0}[/tex] and [tex]v_{x_0}[/tex] for that.
 
  • #12
Let's not have an argument about that--absolutely unnecessary. Now it's been clearly stated that the initial height is 0, and that's all that matters. Anyway:

I've managed to solve for theta. You have written four equations, I'll say 1-4 referring to them in the order you've written them (and I'm just saying t, but it's what you've written as tf--I'm just lazy):

I isolated t2 from equation 3. Plugged this into equation 4 (leave the t that's not squared as it is), creating an equation with only t (equation 5).

I then isolated t from equation 2, and plugged this into equation 5. After some rearranging (and knowing that sin(x)/cos(x) = tan(x), but somehow I just assumed you already knew that), I ended up with tan(x)=4H/L.
 
  • #13
Hannisch said:
I isolated t2 from equation 3. Plugged this into equation 4 (leave the t that's not squared as it is), creating an equation with only t (equation 5).

I then isolated t from equation 2, and plugged this into equation 5. After some rearranging (and knowing that sin(x)/cos(x) = tan(x), but somehow I just assumed you already knew that), I ended up with tan(x)=4H/L.

You get a Scooby snack =D Part of the mystery (for me, at least) is solved! XD
 
  • #14
cojaro said:
[tex]y_0[/tex] and [tex]x_0[/tex] were defined. And I'd never let y or x represent velocity components. I use [tex]v_{y_0}[/tex] and [tex]v_{x_0}[/tex] for that.

NOTE: I only use upper case to emphasize key words. It should NOT be interpreted as yelling, as I detest individuals who lack the communications skills and reserve to discuss as civilized human beings rather than insult or yell.

Unfortunately cojaro, the entire physics community does not necessarily employ the use of the same variable types therefore, clarification is the ONLY way to avoid misinterpretation and potential error due to assumptions. Being that I’m attempting to help someone accurately ascertain the values they require, I will not folly by electing to make assumptions. My question was valid for clarification purposes and it could not be misconstrued as offensive, nor insulting, nor was that its intent.

'Yo' was stated as being equal to zero, so yes, it WAS given a numeric value, but what vertical component of the equation was 'Yo' supposed to be representing? Bear in mind that there is more than one vertical component associated with trajectories (vertical height, vertical velocity component, vertical rise time) and since conventions used to represent these values tends to be highly individualistic, accuracy demands clarification. My integrity intact, I continue with the issue at hand...

Now, his teacher presented a far more complicated trajectory scenario than a simple “open air trajectory”, as the projectile must NOT contact the tunnel's ceiling, which is something that typically wouldn’t be an issue outside of the tunnel. For this reason, additional questions need to be answered. For instance:

How close to the ceiling is an acceptable tolerance for the projectile? This tolerance wasn’t provided either and if we make assumptions here, it may be the reason his answer is assessed as incorrect.

This leads us to the next necessary question:

What is the diameter of the projectile?

If we are to come within a fraction of a meter, or half a meter, or a whole meter, etc... of the ceiling, we need to know the diameter of the projectile because it is the center of mass of the projectile that will achieve the vertical height in the projectile's apex thereby making the projectile exceed its apex by the projectile’s radius. So we must compensate launch velocity (u) and possibly trajectory angle accordingly to meet the as yet unspecified tolerances and projectile dimensions.

Yet another question that requires an answer:

Is the projectile spherical or oblong? If it’s spherical, then knowing its radius will suffice, but if it’s oblong, then its length must be known, as it may rotate lengthwise on its axis after being launched, which again requires that we compensate launch velocity (u) and/or trajectory angle to ensure it DOESN’T contact with the ceiling.

I’m a very thorough individual and I'm only being realistic. His teacher DOESN'T want the projectile to touch the ceiling, so his teacher should have provided ALL the additional required information, not just a portion of it.

Yes, I'd likely give his teacher a headache f-a-s-t, but then, I taught Engineering students for many, many years and I made damn certain I provided ALL the necessary information, leaving nothing undefined in any manner. “Explicitly clear” leaves no room for questions, arguments, nor can it be frowned upon for any reason.
 
  • #15
Quite an interesting problem, thank you!
I wrote cos where I should have written sin in the first step, so I had quite a time.
I even did an example to sort out what was wrong, but I choose a 45 degree angle so it looked like all was well! Finally did a second example with 30 degrees and caught it.
I did let t be the half time (time to max height), which simplified the expressions a bit.

In all these projectile questions, I like to write "horizontal" and "vertical" headings, then d=vt for horizontal and v = vi + at as well as d = vit + .tat^2 for the vertical. So my first 3 equations were
L/2 = u*cos(A)*t (1) 0 = u*sin(A) - gt (2) and H = u*sin(A)*t - .tgt^2 (3)
Sorry, I forget how to write theta so just using A for the angle of elevation.
I solved (2) for t and subbed into (1) to get (4).
Also subbed (2) for t into (3) to get (5).
Using (4) and (5) with no t's, I solved one for u, subbed in the other and got the formula for the angle. Similarly to get the formula for u.
 
  • #16
i think the asumption that the projectile in question can be a particle will sufice, that is to say an object that has mass confined to a single point with no volume. I think the more pressing question is what chance is there of the particle passing straight through the roof of the tunnel thus making the initial velocity easy to work out, and how many people are observing the particle and what effect this will have on the particle lol
and surely we have to acount for the change in hight on the particles jurney and the result this has on the force on the particle due to gravity. =]
 
  • #17
Gnosis said:
I’m a very thorough individual and I'm only being realistic. His teacher DOESN'T want the projectile to touch the ceiling, so his teacher should have provided ALL the additional required information, not just a portion of it.

Gnosis, what's provided in Part 1 is all that my textbook gave me (6th Ed. Wiley/Meriam/Kraige "Engineering Mechanics: Dynamics", if you're interested which book.) Personally, I'm not too fond of my textbook. Each section so far is rarely more than 5 pages, then 2 example problems and 30+ practice problems. Ugh.

Yes, I'd likely give his teacher a headache f-a-s-t, but then, I taught Engineering students for many, many years and I made damn certain I provided ALL the necessary information, leaving nothing undefined in any manner. “Explicitly clear” leaves no room for questions, arguments, nor can it be frowned upon for any reason.

I agree, but sometimes we just have to work with what we got. In this case, not much at all =(
 

Related to Dynamics: Projectile Motion velocity equation

1. What is the equation for calculating velocity in projectile motion?

The equation for calculating velocity in projectile motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. How is velocity related to the angle of projection in projectile motion?

The velocity of a projectile is dependent on both the magnitude and direction of the initial velocity, which is determined by the angle of projection. The greater the angle of projection, the greater the initial velocity and therefore the greater the final velocity.

3. What is the difference between initial velocity and final velocity in projectile motion?

Initial velocity is the velocity of a projectile at the moment it is launched, while final velocity is the velocity at any given point during the projectile's flight. The final velocity will change over time due to the effects of gravity and air resistance.

4. Can the velocity equation be used for both horizontal and vertical components of projectile motion?

Yes, the velocity equation can be used for both horizontal and vertical components of projectile motion. The horizontal component will have a constant velocity, while the vertical component will experience a constant acceleration due to gravity.

5. How does air resistance affect the velocity of a projectile?

Air resistance can decrease the velocity of a projectile by exerting a force in the opposite direction of motion. This force increases as the velocity of the projectile increases, eventually reaching a point where it balances out the force of gravity and the projectile reaches its maximum velocity, known as the terminal velocity.

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