Projectile Motion: Determining Launch Angle

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Homework Help Overview

The problem involves projectile motion, specifically determining the launch angle required for a hunter to hit a moose located 500 meters away and 40 meters below the horizontal. The scenario includes a high muzzle velocity and missing sight on the rifle, leading to a need for calculations involving initial velocity and displacement in both horizontal and vertical directions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conversion of muzzle velocity and the setup of equations for projectile motion. There are attempts to create a system of equations to solve for time and angle, with some expressing concern about the complexity of the calculations. Questions arise regarding the handling of trigonometric terms and the algebraic manipulation of the equations.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for algebraic approaches and discussing potential errors in their calculations. There is recognition of the need for clarity in handling trigonometric identities, and some participants express uncertainty about their current level of understanding in relation to the problem's requirements.

Contextual Notes

Some participants note that they have not yet covered certain trigonometric concepts in their coursework, which may affect their ability to progress with the problem. There is also mention of the complexity of the equations leading to confusion about the expected methods for solving the problem.

EggsBenedict1

Homework Statement



A hunter aims a high velocity semi-automatic rifle at a moose that is 500.0 m away in a direct line of sight. The moose stands 40.0 m below the horizontal in a depression with a bog at the bottom. The muzzle velocity is 2500.0 feet per second and the sight on the rifle is missing.

Determine the angle of launch above the horizontal to hit the moose in the neck that is 2.0 m above the ground where he stands.

Vi=762m/s
dx=500m
dy=-38m

Homework Equations



dy=Viyt+1/2at^2
Vx=dx/t

The Attempt at a Solution



I know that the muzzle velocity of 2500 ft/s needs to be converted to 762m/s.

I've looked at my equation sheet and there are always two or more unknown variables. Because of that, I've tried creating a system of equations. I've rearranged the x velocity equation (Vx=dx/t) to get t=500/762Cos(theta) and substituted that into dy=vit+1/2at^2. However, that turned out to be very complicated and I don't think we'd be expected to do that. I'm assuming I'm over complicating things.
 
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EggsBenedict1 said:

Homework Statement



A hunter aims a high velocity semi-automatic rifle at a moose that is 500.0 m away in a direct line of sight. The moose stands 40.0 m below the horizontal in a depression with a bog at the bottom. The muzzle velocity is 2500.0 feet per second and the sight on the rifle is missing.

Determine the angle of launch above the horizontal to hit the moose in the neck that is 2.0 m above the ground where he stands.

Vi=762m/s
dx=500m
dy=-38m

Homework Equations



dy=Viyt+1/2at^2
Vx=dx/t

The Attempt at a Solution



I know that the muzzle velocity of 2500 ft/s needs to be converted to 762m/s.

I've looked at my equation sheet and there are always two or more unknown variables. Because of that, I've tried creating a system of equations. I've rearranged the x velocity equation (Vx=dx/t) to get t=500/762Cos(theta) and substituted that into dy=vit+1/2at^2. However, that turned out to be very complicated and I don't think we'd be expected to do that. I'm assuming I'm over complicating things.
Welcome to the PF. :smile:

That looks like the correct approach. Can you show us the next steps in your solution for the y(t) values, and show us what time you end up with?
 
berkeman said:
Welcome to the PF. :smile:

That looks like the correct approach. Can you show us the next steps in your solution for the y(t) values, and show us what time you end up with?

Thanks!
-38=(762Sin)(500/762cos)+ (-4.9)(500/762cos)^2

-38=500(sin/cos) + (-4.9)(500^2/762^2)(cos^2)

-30-(500)(sin/cos)=(-2.109)(cos^2)

14.21-(-236.99)(Sin/cos)=cos^2

I don't know how to isolate a variable after this. I know Sin/cos is tan but I don't think that helps.
 
EggsBenedict1 said:
Thanks!
-38=(762Sin)(500/762cos)+ (-4.9)(500/762cos)^2

-38=500(sin/cos) + (-4.9)(500^2/762^2)(cos^2)

-30-(500)(sin/cos)=(-2.109)(cos^2)
O
14.21-(-236.99)(Sin/cos)=cos^2

I don't know how to isolate a variable after this. I know Sin/cos is tan but I don't think that helps.
There are many advantages in working entirely algebraically, not plugging in any numbers until the final step.
Something went wrong in your handling of the cos2 term. Pay close attention to the parentheses.

When you have corrected that error, you will need to use several trig facts. As you say, sin/cos=tan; also 1/cos=sec and sec2=1+tan2.
 
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haruspex said:
There are many advantages in working entirely algebraically, not plugging in any numbers until the final step.
Something went wrong in your handling of the cos2 term. Pay close attention to the parentheses.

When you have corrected that error, you will need to use several trig facts. As you say, sin/cos=tan; also 1/cos=sec and sec2=1+tan2.

I think I know what I did wrong with the parentheses. However, we've never learned about the other trig facts such as sec. So I'm thinking there's either another way to do it or my teacher made a mistake and assumed we were more advanced in pre-calculus than we are. (I'm only about two months into grade 12)
 
EggsBenedict1 said:
I know what I did wrong with the parentheses
So what algebraic equation do you get now, in terms of v, g, y and θ?
EggsBenedict1 said:
I'm thinking there's either another way to do it
You know sin2+cos2=1, right? So (sin/cos)2+1=(1/cos)2.
 

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