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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- Oct 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

- Oct 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 - Oct 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 $\displaystyle v_0 = 1200~ft/s$ and let $\displaystyle \theta$ be the angle of elevation and let t be the time of flight of the bullet.

We know that

$\displaystyle x_0 = 0~ft$ and $\displaystyle y_0 = 0~ft$

$\displaystyle x = 3000~ft$ and $\displaystyle y = 0~ft$

$\displaystyle v_{0x} = 1200~cos(\theta)$ and $\displaystyle v_{0y} = 1200~sin(\theta)$

$\displaystyle a_x = 0~ft/s^2$ and $\displaystyle a_y = -32 ~ft/s^2$

So:

$\displaystyle x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

$\displaystyle 3000 = 1200~cos(\theta)t$

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

$\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$\displaystyle 0 = 1200~sin(\theta)t - 16t^2$

So, solve the y equation for $\displaystyle sin(\theta)$:

$\displaystyle sin(\theta) = \frac{16t}{1200}$

Solve the x equation for $\displaystyle cos(\theta)$:

$\displaystyle cos(\theta) = \frac{3000}{1200t}$

Now use

$\displaystyle sin^2(\theta) + cos^2(\theta) = 1$

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 - Oct 3rd 2007, 04:48 AMtopsquark
- Oct 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.

. . $\displaystyle x \:=\: (v_o\sin\theta)t\qquad y \:=\: h_o + (v_o\sin\theta)t - 16t^2$

where $\displaystyle v_o$ is the initial speed and $\displaystyle \theta$ is the angle of elevation.

We have: $\displaystyle v_o = 1200,\;h_o = 0$

Our equations are: .$\displaystyle \begin{Bmatrix}x \:=\:(1200\cos\theta)t & [1] \\ y\:=\:(1200\sin\theta)t - 16t^2 & [2]\end{Bmatrix}$

Assuming the target is at ground level, we want: .$\displaystyle x = 3000,\;y = 0$

Equation [1] becomes: .$\displaystyle (1200\cos\theta)t \:=\:3000\quad\Rightarrow\quad t \:=\:\frac{5}{2\cos\theta} $

Substitute into $\displaystyle y = 0\!:\;\;(1200\sin\theta)\left(\frac{5}{2\cos\theta }\right) - 16\left(\frac{5}{2\cos\theta}\right)^2\:=\:0$

This simplifies to: .$\displaystyle 3000\tan\theta - 400\sec^2\theta \:=\:0$

Divide by 200: .$\displaystyle 15\tan\theta - 2\sec^2\theta\:=\:0 \quad\Rightarrow\quad15\tan\theta - 2(\tan^2\!\theta + 1) \:=\:0$

We have a quadratic: .$\displaystyle 2\tan^2\!\theta - 15\tan\theta + 2 \:=\:0$

Quadratic Formula: .$\displaystyle \tan\theta \;=\;\frac{15 \pm\sqrt{15^2 - 4(2)(2)}}{2(2)} \;=\;\frac{15 \pm\sqrt{209}}{4}$

And we have two answers:

. . $\displaystyle \begin{array}{ccc}\tan\theta \:=\:\frac{15 + \sqrt{209}}{4} \:=\:7.364208074 & \Rightarrow & \theta \:\approx\:82.27^o \\

\tan\theta \:=\:\frac{15 - \sqrt{209}}{4} \:=\:0.135791926 & \Rightarrow & \theta \:\approx\:7.73^o \end{array} $

The maximum elevation is 82.27°.

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