Find unknown angle of Projectile

In summary: Rightarrow \ t = \frac{y}{vtsin\theta}-1/2gt^2##Now, substitute these into the equation for y:##y=\frac{v}{vtsin\theta}-1/2gt^2##which simplifies to:##y=\frac{v}{vcos\theta}##
  • #1
h.s
6
0
I don't need the solution solved for me, but I need help with the method to find the solution.
1. Homework Statement

A projectile is launched from an unknown angle and expected to hit a target at a lower altitude.
The values given are the initial velocity, and the distance of the target from the launcher.

Homework Equations

The Attempt at a Solution


I've tried solving for time, angle, or final velocity but they each need another unknown variable. Any insight would be helpful.[/B]
 
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  • #2
A trajectory can be described as a parabola of the form ##y = ax^2 + bx + c##, where usually ##c = 0## and ##a, b## are functions of ##v, \theta##.

If you derive this equation that would be a good start.
 
  • #3
i don't have θ. would a be the initial velocity or average velocity? because I can't solve for average velocity
 
  • #4
The parabolic equation is a general equation that doesn't require you to know the numerical value of any parameter. Do you understand the concept of working with unknowns as parameters, such as ##v, \theta, x \ \ and \ \ y##?
 
  • #5
I'm still at a loss. Suppose I apply this equation. The line would pass through two coordinates (0, 1.13) at launch, then at (4.382, 0) when it hits the ground. How would I solve for θ?
 
  • #6
You'll have one equation with one unknown. Solving for ##\theta## should be relatively simple if you can do the maths to get there.
 
  • #7
... I'll have one equation with two unknowns; I still don't have a and b..
 
  • #8
h.s said:
... I'll have one equation with two unknowns; I still don't have a and b..
a and b will be functions of v (which you know) and ##\theta##

I'll give you a start:

##x = vtcos \theta##

##y = ?##

Then, try to eliminate t. It's over to you to do some maths, I'm afraid!
 
  • #9
PeroK said:
a and b will be functions of v (which you know) and θ\theta
What does that mean?
 
  • #10
i know x=vtcosθ and y=vtsinθ-1/2gt^2, but my problem is that I don't know how to go further without knowing t, nor θ
 
  • #11
h.s said:
i know x=vtcosθ and y=vtsinθ-1/2gt^2, but my problem is that I don't know how to go further without knowing t, nor θ

##x=vtcos\theta \ \Rightarrow \ t = \frac{x}{vcos\theta}##
 

1. How do you calculate the unknown angle of a projectile?

To calculate the unknown angle of a projectile, you can use the equation tan(θ) = (vy0 + 1/2gt^2)/vx0, where θ is the unknown angle, vy0 is the initial vertical velocity, g is the acceleration due to gravity, t is the time the projectile is in the air, and vx0 is the initial horizontal velocity.

2. What information do I need to find the unknown angle of a projectile?

In order to find the unknown angle of a projectile, you will need to know the initial vertical and horizontal velocities of the projectile, the acceleration due to gravity, and the time the projectile is in the air.

3. Can I use this equation for any type of projectile?

Yes, this equation can be used for any type of projectile as long as you have the necessary information. However, keep in mind that this equation assumes no air resistance or other external forces acting on the projectile.

4. What units should I use when plugging in values for this equation?

It is important to use consistent units when using this equation. For the velocities, use meters per second (m/s) and for time, use seconds (s). The acceleration due to gravity should be in meters per second squared (m/s^2).

5. Is there a simpler way to find the unknown angle of a projectile?

There are alternative methods for finding the unknown angle of a projectile, such as using trigonometric ratios or vector components. However, the equation provided is a straightforward and reliable method for solving for the unknown angle.

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