Projectile Motion: Determining Launch Angle

[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.

 

[berkeman]

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.

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?

 

[EggsBenedict1]

Welcome to the PF.

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.

 

[haruspex]

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.

 

[EggsBenedict1]

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)

 

[haruspex]

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?

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.