A Projectile Fired At A 45 Degree Angle

In summary: I'm talking about. In summary, the projectile was fired at a 45 degree angle and just barely made it above a 6m high fence that was 100 meters away. Its initial velocity was V0.
  • #1
miniradman
196
0

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 learned 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
miniradman said:

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 learned 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
step 1. Write down all of the equations you know that might be useful here.
 
  • #4
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 don't 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
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
NascentOxygen said:
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
miniradman said:
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
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
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
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
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
I think so, but I don't see how you can get numbers out of those letters :biggrin: :confused:
 
  • #13
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
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
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
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
Cipherflak said:
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?
 
  • #18
miniradman said:
So simultaneous equations?

Yes.
 
  • #19
Cipherflak said:
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
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
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
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
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
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
miniradman said:
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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  • #26
NascentOxygen said:
If you launch an identical projectile from 6 metres above the ground, with the same firing angle (45 deg), at 100m it will now clear a fence 12 metres high.
Touch'e :approve:

Ok, I'm lost... No matter what I do to the simultaneous equations. I never get the right answer :cry:

This is what I did

6=-9.81t^2 + y*(sin45 = 0.7071)----------------1
100=y*(cos45 = 0.7071)*t ----------------------2

make y the subject for eqn 2.

y=100/0.7071t

Sub y value into eqn 1

-6=-9.8t^2*(100/0.7071t)*0.7071

"0.7071" a "t" from -9.8t^t cancel out because of "0.7071" under the "100"

-6=-9.8t+100
100+6=9.8t
106/9.8 = 9.8t/9.8
9.24=t

I know I MUST have made a mistake but the worst part is that I can't see it :cry:
 
  • #27
Rayquesto said:
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?
hypothetically... if the y component was 0 at 100m, it will be at its peak height or on the ground. If it was at its peak height, this means that the time it takes to get there is affected by gravity. The only problem is that 100m mark isn't at peak height or on the ground.
 
  • #28
-6=-9.8t^2*(100/0.7071t)*0.7071

Shouldn't the * be a + ?
 
  • #29
yup, you switched a + for a *. also, it seems you added a minus to the 6. Drop the minus.

(I hope it wasn't because you misread my last post, the -6m thing was just an answer to your hypotethical situation where you stand 6m above ground and just clear a "fence 0 meters high".)
 
  • #30
I'll try it that way.

6=9.81t^2 + y*(sin45 = 0.7071)----------------1
100=y*(cos45 = 0.7071)*t ----------------------2

make y the subject for eqn 2.

y=100/0.7071t

Sub y value into eqn 1

6=9.8t^2*(100/0.7071t)*0.7071

"0.7071" a "t" from -9.8t^t cancel out because of "0.7071" under the "100"

6=9.8t+100
100-6=9.8t
94/9.8 = 9.8t/9.8

9.59=t

My answer should be around 4-5ish seconds
 
  • #31
6=9.8t^2*(100/0.7071t)*0.7071

we just told you that you should have a plus in there, not a multiplication!

you do something wrong when you insert the substitution.
 
  • #32
Ah, found the problem,

s = ut+(1/2) at^2

I used this formula and I got the right answer... thanks anyway guys
 
  • #33
well you can't solve this problem with only one equation.
 
  • #34
Yep, I used that s=ut+1/2 at^2 for both the horizontal and vertical components ;)
 
  • #35
With 0 acceleration in the horionztal I take it. That's actually what I said in my equations.
 
<h2>1. How does the angle of a projectile affect its trajectory?</h2><p>The angle at which a projectile is fired greatly influences its trajectory. At a 45 degree angle, the projectile will follow a parabolic path, reaching the same height as it was launched and landing at the same horizontal distance as it was fired.</p><h2>2. What is the optimal angle for maximum distance when firing a projectile?</h2><p>The optimal angle for maximum distance when firing a projectile is 45 degrees. This angle allows for the longest horizontal distance traveled while also taking into account air resistance and other factors.</p><h2>3. How does air resistance affect a projectile fired at a 45 degree angle?</h2><p>Air resistance can decrease the horizontal distance traveled by a projectile fired at a 45 degree angle. This is due to the fact that air resistance acts in the opposite direction of the projectile's motion, slowing it down and reducing its overall distance traveled.</p><h2>4. Can a projectile fired at a 45 degree angle reach a higher height than its initial position?</h2><p>Yes, a projectile fired at a 45 degree angle can reach a higher height than its initial position. This is because the projectile's initial velocity has both a horizontal and vertical component, allowing it to reach a greater height before falling back to the same horizontal distance as it was fired.</p><h2>5. How does the mass of a projectile affect its trajectory when fired at a 45 degree angle?</h2><p>The mass of a projectile does not significantly affect its trajectory when fired at a 45 degree angle. As long as the initial velocity and angle remain the same, the trajectory will be the same regardless of the projectile's mass. However, a heavier projectile may experience slightly more air resistance, causing it to travel a slightly shorter distance.</p>

1. How does the angle of a projectile affect its trajectory?

The angle at which a projectile is fired greatly influences its trajectory. At a 45 degree angle, the projectile will follow a parabolic path, reaching the same height as it was launched and landing at the same horizontal distance as it was fired.

2. What is the optimal angle for maximum distance when firing a projectile?

The optimal angle for maximum distance when firing a projectile is 45 degrees. This angle allows for the longest horizontal distance traveled while also taking into account air resistance and other factors.

3. How does air resistance affect a projectile fired at a 45 degree angle?

Air resistance can decrease the horizontal distance traveled by a projectile fired at a 45 degree angle. This is due to the fact that air resistance acts in the opposite direction of the projectile's motion, slowing it down and reducing its overall distance traveled.

4. Can a projectile fired at a 45 degree angle reach a higher height than its initial position?

Yes, a projectile fired at a 45 degree angle can reach a higher height than its initial position. This is because the projectile's initial velocity has both a horizontal and vertical component, allowing it to reach a greater height before falling back to the same horizontal distance as it was fired.

5. How does the mass of a projectile affect its trajectory when fired at a 45 degree angle?

The mass of a projectile does not significantly affect its trajectory when fired at a 45 degree angle. As long as the initial velocity and angle remain the same, the trajectory will be the same regardless of the projectile's mass. However, a heavier projectile may experience slightly more air resistance, causing it to travel a slightly shorter distance.

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