Question

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 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:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/$s^{2}$

Bring out the given parameters from the question:

Initial Velocity ($V_{1}$) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$tcos ∅     .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

           t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$g$t^{2}$        ................... Equation 2

Where   $V_{1y}$ = Velocity in the Y- direction

              t  = Time taken

$V_{1y}$ = $V_{1}$sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/($V_{1x}$cos ∅)

                      t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$  =    $\frac{d}{cos o}$             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$g$\frac{2}{cos0} ^{2}$

                                  Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$g$\frac{2}{cos0} ^{2}$

  Vertical Distance = h = dtan∅   - $\frac{1}{2}$g$\frac{2}{cos0} ^{2}$

  Applying geometry

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

  Vertical Distance = h = d tan∅   - 2 g ($tan^{2} o$ + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

              800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6$Q^{2}$ + 19.6

Rearranging 19.6$Q^{2}$ - 2000 Q + 780.4 = 0

                    $Q_{1}$ = 101.6291

                      $Q_{2}$ = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°