Question

A cannon with a muzzle speed of 1003 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 804 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

Answer 1

let the cannon is fired at an angle p and in time t the cannon reached its target

horizontal component of velocity = 1003 cosp
and vertical component = 1003 sinp

as it took time t to reach target
1900 = 1003 cosp t and 804 = 1003 sinp t -1/2 gt²
eliminating t we get
   804 = 1900 tanp-18.56 sec²p = 1900 tanp-18.56-18.56 tan²p
=> 18.56 tan²p-1900 tanp-822.5 = 0
solving for tanp we get tanp = 102.8 and -0.43
as angle should be above horizontal negative value of tan is neglected
therefore tanp=102.3 => cosp = 0.0097

substituing the value in 1st eq. we get t = 195.3 s
and angle is 89.4 degrees