A Projectile Fired At A 45 Degree Angle

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Homework Statement


A projectile is fired at a 45 degree angle and its just barely able to make it above a 6m high fence that is 100 meters away. What was the projectile's initial velocity?


Homework Equations


x and y components?


The Attempt at a Solution


Well the problem with this one is that I don't know where to start. I mean this is basically the reverse of everything I've learnt about projectile motion. I think I should be calculating the x and y components first but I'm not even sure how to do that.
 

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  • #2
Hootenanny
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Homework Statement


A projectile is fired at a 45 degree angle and its just barely able to make it above a 6m high fence that is 100 meters away. What was the projectile's initial velocity?


Homework Equations


x and y components?


The Attempt at a Solution


Well the problem with this one is that I don't know where to start. I mean this is basically the reverse of everything I've learnt about projectile motion. I think I should be calculating the x and y components first but I'm not even sure how to do that.
In all these constant acceleration kinematics problems, the best place to start is by writing two lists. The first is a list of everything that you know already. The second, is a list of the things you want to know. You should do this for each component (vertical and horizontal) of the motion.
 
  • #3
NascentOxygen
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step 1. Write down all of the equations you know that might be useful here.
 
  • #4
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1. v=u+at (cannot use because there is no time or final velocity)
2. s= (u+v)/2 x t (cannot use because no time, distance or inital and final velocity)
3. s= ut + 1/2 at^2 (cannot use because there is no initial or final velocities or time or distance)
4. v^2 = u^2 +2as (cannot use this one because we have no inital or terminal velocity or distance)

Others
s=d/t (I'll most likely (definately) use this at the end when I have my horizontal and veritcle components)

I'm sorry, but I honestly dont see how I can just have 1 variable when using any of these kinematics (when solving for the horizontal and veritcle components). Could using simultaneous equations be answer?

What I already know:
Distance from fence to launch point = 100m
Gravitational Acceleration = -9.8m/s or 9.8m/s
Angle of launch = 45 degrees
the x and y components would be equal? :confused:

What I need to know:
x and y components
total distace covered while in the air
time of flight?
inital velocity
 
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  • #5
NascentOxygen
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It's fired with a velocity V at 45 degrees. Write the horizontal component of firing speed, in terms of V. Write the vertical component of firing speed, in terms of V.
 
  • #6
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It's fired with a velocity V at 45 degrees. Write the horizontal firing speed, in terms of V. Write the vertical firing speed, in terms of V.
Sorry mate, but I'm not exactly sure what is meant by "in terms of" :confused:
 
  • #7
NascentOxygen
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Sorry mate, but I'm not exactly sure what is meant by "in terms of" :confused:

Okay. :smile:

It's fired with a velocity V at 45 degrees. Write the horizontal component of that.

Now write the vertical component.
 
  • #8
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So the verticle component would be

Sinθ = o/V
Sin(45 = o/V


The horizontal component

Cosθ = a/V
Cos(45 = a/V

where hypotenues is equal to V

what now? this is where I get stuck because I don't know if I have enough infomation to move on or am I missing something:confused:
 
  • #9
NascentOxygen
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What is o? What is a? This topic reserves a for acceleration. If you use the same symbols for different things, you will soon get confused.

Besides, I can't see your expression for the horizontal velocity.
 
  • #10
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sorry, I was using trig functions where o = opposite and a = adjacent.

I might just keep it as: a = adj and o = opp

Can I use Cosθ = adj/hyp or speed=distance/time to figure out the horizontal component?
 
  • #11
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if the initial speed is v0, the horizontal component is u0=v0*cosθ and the vertical w0=v0*sinθ.

Do you see this? :)
 
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  • #12
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I think so, but I don't see how you can get numbers out of those letters :biggrin: :confused:
 
  • #13
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one step at a time ;)
you may be used to other notations like v_x instead of u or whatever, but get comfortable with what components mean first.
 
  • #14
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this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that;

6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine)

100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine)


everybody concur? :)
Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
 
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  • #15
NascentOxygen
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Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
Yes, it does look like 3 unknowns, but since we were told that theta is 45 degrees, then that leaves only two unknowns.
 
  • #16
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You're absolutely right, I forgot to insert for theta. It's best to keep the symbols as long as possible though, to obtain a more general solution :)
 
  • #17
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this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that;

6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine)

100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine)


everybody concur? :)
Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
So simultaneous equations?
 
  • #19
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this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that;

6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine)

100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine)


everybody concur? :)
Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
I'm not sure if this is correct, but what if I say that the launch point was 6 metres above ground and I removed the fence. Would the time of flight be that same if the launch was on the ground and the fence was 100m away?
 
  • #20
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if the velocity of the y component is zero at 100 meters away and Vy=V0 - gt, then what does that tell you about the time it takes to get there?
 
  • #21
NascentOxygen
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How would you define the path that the parabola must follow, then, with no obstacle for it to clear? Try a few sketches.

Have you finished solving the problem at hand?
 
  • #22
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Not yet, but I have drawn a sketch of this two to show my line of thought.

As for finishing the problem at hand... I didn't really understand how it works (it seems like maths is a world apart). I don't know what V0 or Vy means?
 

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  • #23
NascentOxygen
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If you launch an identical projectile from 6 metres above the ground, with the same firing angle (45 deg) and speed, at 100m it will now clear a fence 12 metres high.
 
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  • #24
NascentOxygen
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I'll summarize how these motion at the earth's surface problems are solved.

Horizontally, with no air resistance, projectiles travel at a constant speed. This means that its shadow travels over the land at a constant speed. What speed would this be? It's the same horizontal component of speed with which it was launched. Even when the projectile soars way up into the sky, before returning, following a high arc, its shadow all the time sweeps across the landscape at a steady speed.

Vertically, the motion is affected by gravity, and the vertical component of velocity follows the equations you know that involve acceleration "a" which is gravity. Gravity slows down the vertical ascent speed until the projectile eventually ceases vertical motion for a moment (i.e., is neither going up nor down), then it speeds up again but now in a downwards direction.
 
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  • #25
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I'm not sure if this is correct, but what if I say that the launch point was 6 metres above ground and I removed the fence. Would the time of flight be that same if the launch was on the ground and the fence was 100m away?

No, my equations are correct.

If you stand on your fence and hit the ground 100 away instead, the bullet would need a different intitial velocity and time of flight. (swap 6m with -6m and work from there).
 
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