muzzle velocity projectile

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October 3rd 2007, 03:03 AM
harry

muzzle 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 AM
salaam786

T-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 AM
topsquark

Quote:

Originally Posted by harry

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

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

We know that
$x_0 = 0~ft$ and $y_0 = 0~ft$
$x = 3000~ft$ and $y = 0~ft$
$v_{0x} = 1200~cos(\theta)$ and $v_{0y} = 1200~sin(\theta)$
$a_x = 0~ft/s^2$ and $a_y = -32 ~ft/s^2$

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

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

This equation has two unknowns in it. So we need another equation with the same two unknowns:
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

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

So, solve the y equation for $sin(\theta)$ :
$sin(\theta) = \frac{16t}{1200}$

Solve the x equation for $cos(\theta)$ :
$cos(\theta) = \frac{3000}{1200t}$

Now use
$sin^2(\theta) + cos^2(\theta) = 1$
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
October 3rd 2007, 04:48 AM
topsquark

Quote:

Originally Posted by salaam786

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

:mad: What does this have to do with harry's question??

-Dan
October 3rd 2007, 08:06 AM
Soroban

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.

We're expected to know the "Projectile Equations":
. . $x \:=\: (v_o\sin\theta)t\qquad y \:=\: h_o + (v_o\sin\theta)t - 16t^2$
where $v_o$ is the initial speed and $\theta$ is the angle of elevation.

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

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

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

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

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

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

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

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

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

And we have two answers:
. . $\begin{array}{ccc}\tan\theta \:=\:\frac{15 + \sqrt{209}}{4} \:=\:7.364208074 & \Rightarrow & \theta \:\approx\:82.27^o \\<br /> \tan\theta \:=\:\frac{15 - \sqrt{209}}{4} \:=\:0.135791926 & \Rightarrow & \theta \:\approx\:7.73^o \end{array}$

The maximum elevation is 82.27°.

Did you notice that the two angles are complements?