A shot fired from a gun with a muzzle velocity of 1200ft per second is to hit a target 3000 ft away. Determine the maximum angle of elevation for the gun

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- October 3rd 2007, 03:03 AMharrymuzzle velocity projectile
A shot fired from a gun with a muzzle velocity of 1200ft per second is to hit a target 3000 ft away. Determine the maximum angle of elevation for the gun

- October 3rd 2007, 03:54 AMsalaam786T-shapes coursework urgent help needed!!!!!
does anyone know howto answer this question about t-shapes for cwk. i have no idea and i would be very grateful if someone could help me with it :confused:

my first question;

My Question Is About T-shapes: Look At A 9 By 9 Grid And Draw T-shapes 3/4, And Compare The T Number (Tn) And T-total (Tt). Can Anyone Help Me?

my second question;

Luk At A 8 By 8, 7 By 7, 6 By 6 And 5 By 5 Grid And Luk At The Relationships Between The T-n And T- Total, And Draw Tables To Show These Relationships, Can You Help Me?

please please try to help me and tell other people on this website so they can help me. thnkyou - October 3rd 2007, 03:58 AMtopsquark
Set up a coordinate system with the origin at the end of the gun and a +x direction in the direction of the target and +y straight up. Let and let be the angle of elevation and let t be the time of flight of the bullet.

We know that

and

and

and

and

So:

This equation has two unknowns in it. So we need another equation with the same two unknowns:

So, solve the y equation for :

Solve the x equation for :

Now use

to write an equation for t.

Why don't you try it from here. There will be two solutions for t. One of these will give a larger angle than the other. Since you are looking for the max angle, this will be the one you want.

-Dan - October 3rd 2007, 04:48 AMtopsquark
- October 3rd 2007, 08:06 AMSoroban
Hello, Harry!

I have a different approach . . .

Quote:

A shot fired from a gun with a muzzle velocity of 1200 ft/sec

is to hit a target 3000 ft away.

Determine the maximum angle of elevation for the gun.

. .

where is the initial speed and is the angle of elevation.

We have:

Our equations are: .

Assuming the target is at ground level, we want: .

Equation [1] becomes: .

Substitute into

This simplifies to: .

Divide by 200: .

We have a quadratic: .

Quadratic Formula: .

And we have two answers:

. .

The maximum elevation is 82.27°.

Did you notice that the two angles are*complements*?