Projectile Motion: Finding Initial Speed for Targeted Height

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In summary: I can do this, it's for fun if you want.. hah yeah right?The bullet will first hit the floor, at x=-9.8 meters.
  • #1
bbhh
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Homework Statement


A gun is located at the origin. The gun is aimed 36.87 degrees above the horizontal. A vertical wall is located at x = 18m. The flat and level floor is at y=0m. Ignore air friction.

a) What is the minimum initial speed necessary for the bullet to reach the wall without hitting the floor?

b) Suppose the initial speed is 15 m/s. Does the bullet first hit the floor or the Wall? Where does it hit? (I can do this part, just thought I'd throw it in for some reason.)

c) What initial speed is required in order to hit the wall 3.00m above the floor?

Homework Equations


d=vt
Vf^2 = Vo^2 +ad
d = Vot + 1/2at^2

The Attempt at a Solution



a)

I'm really having troubles with this.

I don't know the initial or final velocity or time. I thought that since that since the bullet will hit the wall at the same height that it was fired from, then half way along (x=9m) the bullet would peak, so I could use that point knowing the Final velocity in the y would be 0... then I could solve for the initial velocity in the Y, then use trig to find the initial velocity.

Vfy^2 = Voy^2 + 2adthen: d = v*(2)t

but unfortunately I don't know what the displacement (in the y) would be at this point.

I'm pretty sure I need to find the initial velocity in the Y, and then time before I can do anything else, but I'm totally stuck.

I might be on the complete wrong track

b)

I can do this, it's for fun if you want.. hah yeah right?

c)

this is basically the same as a) i think...

Please help, my final is tomorrow and this is the only type of projection question I can't figure out.
 
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  • #2
a)

A way to solve any projectile question is to first list theses 6 equations:

[tex] a_{x} = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ a_{y} = -g [/tex]
[tex]v_{xf} = v_{i}cos \theta \ \ \ \ \ \ \ \ \ \ v_{yf} = v_{i}sin \theta - g*t [/tex]
[tex]d_{x} = v_{i}cos \theta * t \ \ \ \ \ \ \ H = v_{i}sin \theta* t - 1/2*g*t^2 [/tex]

In this case you only need the last two equations. (Most often).

So you have 2 equations with 2 unknowns( t and vi). Then you can solve for.
 
  • #3
okay, thanks for your help. that doesn't really get me much farther though, I'm pretty sure those are a few of the equations i had but with the trig worked into them, can you show me the next step? like I said, I don't know vi or t, as far as i can tell...
 
  • #4
You know the horizontal distance travelled, so find an equation involving this distance. It will be in terms of t and V_0. So now you can get an expression for the time at which the particle hits the wall in terms of V_0. Now, what can you say about the vertical distance traveled when it hits the wall, specifically the case of the least velocity needed?
 
  • #5
d = Vo(cosTheta)*t
Vo(cosTheta) = d/t

I don't know the vertical distance, the vertical displacement is 0.
 
  • #6
Exactly, so now you have enough figured out to solve it, combine d_x with d_y.
 
  • #8
nevermind! i got it. wooooo. thanks for your help!
 

1. How is the trajectory of a projectile affected by air resistance?

The trajectory of a projectile is affected by air resistance because it creates a force in the opposite direction of the projectile's motion. This force, known as drag, slows down the projectile and changes its trajectory, causing it to fall at a steeper angle.

2. What factors influence the range of a projectile?

The range of a projectile is influenced by several factors, including the initial velocity, angle of launch, air resistance, and gravitational pull. Other factors such as wind and altitude can also have an impact on the range of a projectile.

3. How does the mass of a projectile affect its motion?

The mass of a projectile does not affect its motion in terms of horizontal velocity and range. However, a heavier projectile will experience less air resistance and will therefore travel further than a lighter projectile launched at the same velocity.

4. What is the maximum height reached by a projectile?

The maximum height reached by a projectile is determined by the initial velocity and angle of launch. The maximum height occurs at the apex of the projectile's trajectory, where its vertical velocity becomes zero before accelerating back towards the ground.

5. How does the angle of launch affect the motion of a projectile?

The angle of launch has a significant impact on the motion of a projectile. A projectile launched at a lower angle will have a longer range but a lower maximum height, while a projectile launched at a higher angle will have a shorter range but a higher maximum height.

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