Question

A mortar crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of θ = 62.0 degrees, the crew fires the shell at a muzzle velocity of 149 feet per second. How far down the hill does the shell strike if the hill subtends an angle φ = 34.0 degrees from the horizontal? (Ignore air friction.)How long will the mortar shell remain in the air?How fast will the shell be traveling when it hits the ground?

Answer 1

0 = (149 sin62) t + (-32.174 cos34) t^2 /2

t = 9.86 sec ......Ans (time in air)

distance travelled along the hill,

x = (149 cos62)(9.86) + (32.174 sin34)(9.86^2)/2

x = 1564.3 ft .....Ans

vx = (149 cos62) + (32.174 sin34)(9.86)

vx = 247.3 ft/s

vy = (149 sin62) + (-32.174 cos34)(9.86)

vy = - 131.44 ft/s

Speed = sqrt(vx^2 + vy^2) = 280 ft/s .....Ans