Motion in Three-Space

**jffyx** · March 10th 2009, 12:20 AM

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use

as the acceleration of gravity.
Answer in radians.

Thanks!

**mr fantastic** · March 10th 2009, 12:54 AM

Originally Posted by jffyx

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use

as the acceleration of gravity.
Answer in radians.

Thanks!

It will depend on what height you want to hit the object at. Or is it to be assumed that the object is at ground level ....?

**jffyx** · March 10th 2009, 01:46 AM

Originally Posted by mr fantastic

It will depend on what height you want to hit the object at. Or is it to be assumed that the object is at ground level ....?

Oh. It is to be assumed that the object is at ground level.

**mr fantastic** · March 10th 2009, 04:51 AM

Originally Posted by jffyx

Oh. It is to be assumed that the object is at ground level.

Substitute the data into equation (7) here: Projectile Motion

Now solve this equation for $\theta$ .

**Soroban** · March 10th 2009, 08:00 AM

Hello, jffyx!

A gun has a muzzle speed of 90 meters per second.
What angle of elevation should be used to hit an object 150 meters away?

The equations for projectile motion are: . $\begin{array}{ccc}x \;=\; (v_o\cos\theta)t \\ y \;=\; h_o + (v_o\sin\theta)t - 4.9t^2 \end{array}$
. . where: . $\begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}$

We are given: . $h_o = 0,\;v_o=90$

The equations are: . $\begin{array}{cc}x \;=\;(90\cos\theta)t & {\color{blue}[1]} \\ y \;=\;(90\sin\theta)t - 4.9t^2 & {\color{blue}[2]} \end{array}$

Since $x = 150$ , [1] becomes: . $(90\cos\theta)t \:=\:150 \quad\Rightarrow\quad t \:=\:\frac{5}{3\cos\theta}$

Substitute into $y = 0\!:\;\;(90\sin\theta)\left(\frac{5}{3\cos\theta}\ right) - 4.9\left(\frac{5}{3\cos\theta}\right)^2 \:=\:0$

which simplifies to: . $1350\tan\theta - 122.5\sec^2\!\theta \:=\:0$

. . $1350\tan\theta - 122.5(\tan^2\!\theta + 1) \:=\:0 \quad\Rightarrow\quad 122.5\tan^2\!\theta - 1350\tan\theta + 122.5 \:=\:0$

Quadratic Formula: . $\tan\theta \;=\;\frac{1350 \pm\sqrt{1,\!762,\!475}}{245} \;=\;\begin{Bmatrix}0.091500952 \\ 10.92890771 \end{Bmatrix}$

. . $\tan\theta \,=\,0.091500952\quad\Rightarrow\quad \theta \:\approx\:5.228^o$

**Mentia** · March 10th 2009, 09:06 AM

Note how Soroban got two solutions. You could also fire the gun at about 84.77 degrees and hit the target.

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