A mortar* crew is positioned near the top of a steep hill. Enemy forces are char

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 theta = 53.0degree (as shown), the crew fires the shell at a muzzle velocity of 184 feet per second. How far down the hill does the shell strike if the hill subtends an angle Phi = 33.0degree from the horizontal? (Ignore air friction.) How long will the mortar shell remain in the air? How fast will the shell be traveling click to edit he ground?

Explanation / Answer

184 ft/sec= 184*0.3048 = 56.08 m/s

vertical distance travelled is y = d*sin(phi)

now using -y = (uy*t)-(0.5*g*t^2) = (u*sin(theta)*t)-(0.5*9.81*t^2)

-d*sin(phi) = (56.08*sin(53)*t)-(0.5*9.81*t^2)............(1)

horizontal distance travelled is

d*cos(phi) = u*cos(theta)*t

d*cos(33) = 56.08*cos(53)*t.............(2)

(1)/(2)

-tan(33) = ((56.08*sin(53)*t)-(0.5*9.81*t^2)) / (56.08*cos(53)*t)

-0.65 = ((44.78)-(4.905*t))/33.74

-21.93 = 44.78-4.905*t

t = 13.6 s

then from eqn(1) we get

-d*sin(33) = (56.08*sin(53)*13.6)-(0.5*9.81*13.6^2) = -298.11

d = 547.37 m

upto t = 13.6 s the shell remains in air

vx = (u*cos(theta)) = (56.08*cos(53)) = 33.74 m/s

vy = u*sin(53) - (g*t)

vy = (56.08*sin(53))-(9.81*13.6) = -88.62 m/s

so the answer velocity of shell at the ground is sqrt(33.74^2+88.62^2) = 94.82 m/s

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