# Thread: muzzle velocity projectile

1. ## 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

2. ## 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
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

3. 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 $\displaystyle v_0 = 1200~ft/s$ 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

4. 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
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
What does this have to do with harry's question??

-Dan

5. Hello, Harry!

I have a different approach . . .

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