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

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/

Bring out the given parameters from the question:

Initial Velocity () = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = tcos ∅     .......................... Equation 1

where = velocity in the X - direction

           t = Time taken

Vertical Distance = y = t - g        ................... Equation 2

Where   = Velocity in the Y- direction

              t  = Time taken

= sin∅

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

                    t = d/(cos ∅)

                      t = =  =                 ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = - g

                                  Vertical Distance = h = sin∅ - g

  Vertical Distance = h = dtan∅   - g

  Applying geometry

                              = + 1

  Vertical Distance = h = d tan∅   - 2 g ( + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( + 1)

              800 = 2000 tan ∅ - 19.6( + 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 + 19.6

Rearranging 19.6 - 2000 Q + 780.4 = 0

                    = 101.6291

                      = 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°

the answer is to check all of them! they all apply well at least i think so!