I used to be good at phyics, but

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I used to be good at phyics, but....

That was 1987!!! I was at the range the other day with some guys when the question of maximum cannon range came up. I knew it had to be 45 degrees but that was all I could remember.

Homework Statement



A cannon fires a 1kg iron ball at a 45 degree angle with an initial muzzle velocity of 500 m/sec and a muzzle height of 5 ft. Assume level terrain, no wind resistance, and no other complicating factors. Find the time in flight, maximum height, and total horizontal distance travelled.


Homework Equations



All I remember is that 9.8m/s^2 is part of this

The Attempt at a Solution



Hide in corner in fetal position remembering physics class again! :cry: Seriosuly, go to the pros when you want the right answer. Can you please tell me what is the formula for these answers? Thanks!
 

Answers and Replies

  • #2
Hootenanny
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Welcome to Physics Forums.

The first step is to split the velocity into it's vertical and horizontal components. Can you do that?
 
  • #3
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:grumpy: Arrrgh! You are going to make me think?!?!?

Oh, well! It is either the sin or cos of 45. Maybe it's pythagorean's theorem? Is it 353.56 for initial x and y?

(Honestly, I used to get 100% on my physics tests!)
 
  • #4
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Seems ok so far. So you have your inital component velocities.

What do you think now?
 
  • #5
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So we start off with an initial x and y velocity of ~353m/sec; the x component will stay that throughout the time in flight (since we will be ignoring wind resistance and other effects). The Y velocity will start to bleed off at a rate to be determined somewhow by that 9.8 m/s^2 thing, but I have no clue from this point.
 
  • #6
Hootenanny
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So we start off with an initial x and y velocity of ~353m/sec; the x component will stay that throughout the time in flight (since we will be ignoring wind resistance and other effects). The Y velocity will start to bleed off at a rate to be determined somewhow by that 9.8 m/s^2 thing, but I have no clue from this point.
Good. So you know that the distance travelled in the x-direction,x, at time, t, is given by,

x = vxt

where vx is the component of the velocity in the x-direction. Agreed?

Now for the y-direction, the cannon ball undergoes uniformly accelerated motion (since it is being accelerated towards the earth at g). Can you remember any kinematic (SUVAT) equations?
 
  • #7
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x = vxt

where vx is the component of the velocity in the x-direction. Agreed?
QUOTE]

Agreed. But no I do not remember any SUVAT equations (has physics changed in 20 years?!!)
 
  • #9
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The first four I understand.

The fifth,

x = x_0 + v_0 t + (1/2) a t^2

I believe says that the x-coordinate will be the original starting point (x=0 in this case) plus the original velocity multiplied by time plus 1/2 acceleration multiplied by the time squared.

In this case, I could tell you x only when I know t and that was the answer sought (in addition to distance).

The sixth equation I believe says that the new velocity will be the old velocity plus a factor of the acceleration that was possible over the change in distance in question. The fourth equation seems vaguely related to the 6th.

I need advil!
 
  • #10
Hootenanny
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The first four I understand.

The fifth,

x = x_0 + v_0 t + (1/2) a t^2

I believe says that the x-coordinate will be the original starting point (x=0 in this case) plus the original velocity multiplied by time plus 1/2 acceleration multiplied by the time squared.

In this case, I could tell you x only when I know t and that was the answer sought (in addition to distance).

The sixth equation I believe says that the new velocity will be the old velocity plus a factor of the acceleration that was possible over the change in distance in question. The fourth equation seems vaguely related to the 6th.

I need advil!
Good.

So in the previous equation: x = vxt, you know the velocity and you want to find out the horizontal displacement (range), but you know know the time. So, which one of the equations that you just saw would be appropriate to use? Bear in mind that you know the acceleration and the initial and final displacements...
 
  • #11
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Good.

So in the previous equation: x = vxt, you know the velocity and you want to find out the horizontal displacement (range), but you know know the time. So, which one of the equations that you just saw would be appropriate to use? Bear in mind that you know the acceleration and the initial and final displacements...
I do not know the time. Or the initial and final displacments.
 
  • #12
Hootenanny
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I do not know the time.
Indeed you do not, but you can work it out using one of these formulae.
Or the initial and final displacments.
Yes you do, remember that you are trying to find the range of the projectile.
 
  • #13
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Yes you do, remember that you are trying to find the range of the projectile.
Yes, I am trying to find the range, which would be x. The original x=0 (the muzzle of the cannon), the original y=2 (height of muzzle... i used 5 feet, let's just call it 2 meters). The final x (the displacement) I do not know.
 
  • #14
Hootenanny
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Yes, I am trying to find the range, which would be x. The original x=0 (the muzzle of the cannon), the original y=2 (height of muzzle... i used 5 feet, let's just call it 2 meters). The final x (the displacement) I do not know.
Yes, but you missed out one important bit of information that you do know: the final vertical displacement of the cannon ball.
 
  • #15
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The final vertical displacement would be -2meters. Am I missing something? I have no idea ho high it went nor the time in flight (both of which would be related to 9.8m/s^2) nor the distance travelled (which would be dependent on the time in flight).

What am I mssing here?
 
  • #16
Hootenanny
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The final vertical displacement would be -2meters. Am I missing something? I have no idea ho high it went nor the time in flight (both of which would be related to 9.8m/s^2) nor the distance travelled (which would be dependent on the time in flight).
You are indeed correct! Can you use the information you have, i.e. y0 = 0, y1 = -2, a=-9.81 and vy0 = 353.56, together with the formula that you quoted earlier,
x = x_0 + v_0 t + (1/2) a t^2
to determine the flight time?
 
  • #17
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You are indeed correct! Can you use the information you have, i.e. y0 = 0, y1 = -2, a=-9.81 and vy0 = 353.56, together with the formula that you quoted earlier,

to determine the flight time?
No, I cannot, because, although I know Xo, Vo, and A, there are 2 unknowns in that equation: X and T.:cry:
 
  • #18
Hootenanny
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No, I cannot, because, although I know Xo, Vo, and A, there are 2 unknowns in that equation: X and T.:cry:
Yes you can! t (the flight time) is the only unknown in the formula:

y1 = y0 + vy 0t + 1/2 ayt^2
 
  • #19
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Yes you can! t (the flight time) is the only unknown in the formula:

y1 = y0 + vy 0t + 1/2 ayt^2
-2=0+353.56+0.5(-9.8)T^2
-4.9T^2=-355.56
T^2=+72.56
T=8.52 sec

x = x_0 + v_0 t + (1/2) a t^2
x=0 + 353.56m/sec*8.52sec + 4.9m/sec^2*72.56sec^2
x=3012.33m+355.54m
x=3367.87m

The Ymax I thought I would figure by dividing the time in half, but being a perfectionist, how would I account for the milliseconds it would take to fall the last two meters below the cannon's muzzle height?

Also, does my X distance result take into account the distance to it contacts the ground or the distance until it crosses the 2m high mark?
 
  • #20
Hootenanny
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-2=0+353.56+0.5(-9.8)T^2
-4.9T^2=-355.56
T^2=+72.56
T=8.52 sec
This is not correct. You seem to be missing another t
The Ymax I thought I would figure by dividing the time in half, but being a perfectionist, how would I account for the milliseconds it would take to fall the last two meters below the cannon's muzzle height?

Also, does my X distance result take into account the distance to it contacts the ground or the distance until it crosses the 2m high mark?
You have already taken the distance between the cannon and the ground into account. The cannon is at y = 0. Therefore, the cannon ball's initial vertical position is y=0. When the cannon ball hits the ground it is 2 meters below it's initial position, in order words, y=-2. Do you see?
 
  • #21
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This is not correct. You seem to be missing another t

QUOTE]

-2=0+353.56*T+0.5(-9.8)T^2
-2=353.56T-4.9T^2
2=4.9T^2-353.56T
2=4.9(T^2-72.16T)
.408=T^2-72.16T
(looking like a quadratic equation..... begin shutdown sequence...:frown:
 
  • #22
Hootenanny
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-2=0+353.56*T+0.5(-9.8)T^2
-2=353.56T-4.9T^2
2=4.9T^2-353.56T
2=4.9(T^2-72.16T)
.408=T^2-72.16T
(looking like a quadratic equation..... begin shutdown sequence...:frown:
Come on, you're not telling me that you can't solve a quadratic equation?!
 
  • #23
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Not anymore! You need an emoticon for "my head is about to burst!"

Well, going online, I found the solutions to be:

-.005653 (disregard)
72.16

So T=72.16

x = x_0 + v_0 t + (1/2) a t^2
x=0 + 353.56m/sec*72.16sec+ 4.9m/sec^2*(72.16sec)^2
x=25512.89m+25514.62m=51027.51m=51.027km=no way!

can you help show me where am I going wrong?
 
  • #24
Hootenanny
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Not anymore! You need an emoticon for "my head is about to burst!"

Well, going online, I found the solutions to be:

-.005653 (disregard)
72.16

So T=72.16
Be careful with your rounding, you're off by a couple of decimal places here. I have 72.09 seconds.
x = x_0 + v_0 t + (1/2) a t^2
x=0 + 353.56m/sec*72.16sec+ 4.9m/sec^2*(72.16sec)^2
What is the acceleration in the x direction?
 
  • #25
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Be careful with your rounding, you're off by a couple of decimal places here. I have 72.09 seconds.

What is the acceleration in the x direction?
zero! :blushing:

x = x_0 + v_0 t + (1/2) a t^2
x=0 + 353.56m/sec*72.09sec+ zero*(72.16sec)^2
x=25488.14m+0=25.488km=seems a little far to me still
 

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