Question

A cannon with a muzzle speed of $\mathbf{1000}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$is used to start an avalanche on a mountain slope. The target is$\mathbf{2000}\mathbf{}\mathit{m}$from the cannon horizontally and$\mathbf{800}\mathbf{}\mathit{m}$above the cannon. At what angle, above the horizontal, should the cannon be fired?

Short answer

The muzzle is fired at $22.{34}^{\circ }$from horizontal.

Step-by-step solution

Given data:

The speed of the muzzle is$1000\text{m/s}$. The horizontal distance of the target is $2000\text{m}$and vertical distance of the target is$800\text{m}$above the cannon.

The acceleration due to gravity is $9.8{\text{m/s}}^2$.

Understanding the concept:

The expression for vertical distance, from the kinematic law of motion is,

$\mathit{y}\mathbf{=}{\mathit{v}}_{\mathbf{y}}\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}}$

Here,$\mathit{g}$is the acceleration due to gravity.

The expression for the velocity in horizontal direction is,

${\mathit{v}}_{\mathbf{x}}\mathbf{=}\mathit{v}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

Here,$\mathit{v}$is the speed of the muzzle,$\mathit{\theta }$is the angle of muzzle from horizontal axis.

The expression for the velocity in vertical direction is,

${\mathit{v}}_{\mathbf{y}}\mathbf{=}\mathit{v}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

Determine the angle above the horizontal through which the canon should be fired:

The expression for the velocity in horizontal direction is,

$v_{\mathrm{x}}=v\cos \theta$

The expression for the velocity in vertical direction is,

$v_{\mathrm{y}}=v\sin \theta$

The expression for the horizontal distance is,

$x=v_{\mathrm{x}}\times t$

Substitute $v\cos \theta$for $v_{\mathrm{x}}$in the above expression.

$x=v\cos \theta \times t$

Rearrange the above expression for the value of$t$.

$t=\frac{x}{v\cos \theta }$

The expression for vertical distance, from the kinematic law of motion is,

Substitute$v\sin \theta$for$v_{\mathrm{y}}$and$\frac{x}{v\cos \theta }$for$t$in the above expression.

$\begin{array}{c}y=v\sin \theta \left(\frac{x}{v\cos \theta }\right)-\frac{1}{2}g{\left(\frac{x}{v\cos \theta }\right)}^2\\ =x\tan \theta -\frac{g{x}^2}{2v^2}\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta }\right)\\ =x\tan \theta -\frac{g{x}^2}{2v^2}\left(1+{\tan }^2\theta \right)\end{array}$

Substitute 800 for $y,2000\text{m}$for $x,1000\text{m}$for $v$and $9.81{\text{m/s}}^2$for $g$in the above equation,

role="math" localid="1663621114894" $\begin{array}{c}800=\left(2000\right)\tan \theta -\frac{(9.8){\left(2000\right)}^2}{2{\left(1000\right)}^2}\left(1+{\tan }^2\theta \right)\\ 0=19.6{\tan }^2\theta -2000\tan \theta +819.60\\ \tan \theta =0.411,101.62\\ \theta ={\tan }^{-1}(0.411),{\tan }^{-1}(101.62)\\ \theta =22.34°, 89.43°\end{array}$

Take$22.34°$as the angle's value because$89.43°$is close to$90°$that is not possible.

Therefore, the muzzle is fired at $22.34°$from horizontal.