Question

A cannon with a muzzle speed of 1002 m/s is used to start an avalanche on a mountain slope. The target is 2050 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

Answer 1

distance travelled x = 2050 m

x = vox*T

T = x/vox.......(1)

along vertical

initial velocity voy = vo*sintheta

vertical distance travelled = y = 800 m

acceleration = ay = -9.8 m/s^2

from equation of motion

y = voy*T + 0.5*ay*T^2 ............(2)

from 1 & 2

y = (vo*sintheta*x)/(vo*costheta) - 0.5*g*x^2/(vo^2*cos(thet)^2)

y = (vo^2*x*sintheta*costheta - 0.5*9.8*x^2)/(vo^2*(costheta)^2)

y = (vo^2*x*sqrt(1-(costheta)^2)*costheta - 0.5*9.8*x^2)/(vo^2*(costheta)^2)

800 = (1002^2*2050*sqrt(1-(costheta)^2)*costheta - 0.5*9.8*2050^2)/(1002^2*(costheta)^2)

theta = 21.8 degrees