Question

a cannon with a muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. the target is 2000 m from the cannon horizontally and 800 m above the cannon. at what angle, above the horizontal should the cannon be fired?

Answer 1

Answer on Question #53396, Physics / Mechanics | Kinematics | Dynamics

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

Solution:

Given:

$x_{1} = 2000\mathrm{m}$

$h = 800\mathrm{m}$

$v_{0} = 1000\mathrm{m / s},$

$\theta = ?$

Neglecting air resistance, the projectile is subject to a constant acceleration $g = 9.81 \, \text{m/s}^2$ , due to gravity, which is directed vertically downwards.

Equations related to trajectory motion (projectile motion) are given by

Horizontal distance, $x = v_{0x}t$

Vertical distance, $y = v_{0y}t - \frac{1}{2} gt^2$

where $v_{0}$ is the initial velocity.

From first equation,

$$t _ {1} = \frac {x _ {1}}{v _ {0 x}} = \frac {x _ {1}}{v _ {0} \cos \theta} = \frac {2 0 0 0}{1 0 0 0 \cos \theta} = \frac {2}{\cos \theta}$$

Substituting to second equation,

$$h = v _ {0 y} \frac {x _ {1}}{v _ {0} \cos \theta} - \frac {1}{2} g (\frac {2}{\cos \theta}) ^ {2}$$ $$h = v _ {0} \sin \theta \frac {x _ {1}}{v _ {0} \cos \theta} - \frac {1}{2} g (\frac {2}{\cos \theta}) ^ {2}$$ $$h = x _ {1} \tan \theta - \frac {1}{2} g (\frac {2}{\cos \theta}) ^ {2}$$

Applying trigonometric identities:

$$\frac {1}{\cos^ {2} \theta} = \frac {\sin^ {2} \theta + \cos^ {2} \theta}{\cos^ {2} \theta} = \frac {\sin^ {2} \theta}{\cos^ {2} \theta} + 1 = \tan^ {2} \theta + 1$$

$h=x_{1}\tan\theta-2g\ (\tan^{2}\theta+1)$

$800=2000\tan\theta-19.6\ (\tan^{2}\theta+1)$

Substituting

$z=\tan\theta$

$800=2000z-19.6\ (z^{2}+1)$

$19.6z^{2}-2000z+819.6=0$

Solutions of quadratic equation:

$z_{1}=$ $0.411459$

$z_{2}=$ $101.629$

Back to $\tan(\theta)=z$

$\tan(\theta_{1})=$ $0.411459$

$\tan(\theta_{2})=$ $101.629$

$\theta_{1}=$ $\tan^{-1}(0.411459)=22.37{}^{\circ}$

$\theta_{2}=$ $\tan^{-1}(101.629)=89.44{}^{\circ}$

Answer: $22.37{}^{\circ}$ or $89.44{}^{\circ}$

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