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 = 54.0o (as shown), the crew fires the shell at a muzzle velocity of 204 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 33.0o 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?

Explanation / Answer

along horizantal displacement = x = d*cosphi = d*cos33

initial speed = vox = vo*costheta

x = vox*T

T = x/vox = (d*cosphi)/(vo*costheta)

along vertical

displacement y = -d*sinphi

initial velocity = voy = vo*sintheta

y = voy *T + 0.5*ay*T^2

-d*sinphi = (vo*sintheta*d*cosphi)/(vo*costheta) - 0.5*g*d^2*(cosphi)^2/(vo^2*(costheta)^2)

-d*sinphi = (tantheta*d*cosphi)) - 0.5*g*d^2*(cosphi)^2/(vo^2*(costheta)^2)

-d*sin33 = (d*sin33*tan54)-(0.5*32*d^2*(cos33)^2)/(204^2*(cos54)^2))

d = 1653.6 feet

+++++++++++++++

T = (d*cos33)/(vo*cos54)

T = (1653.6*cos33)/(204*cos54)

T = 11.56 s

++++++++++++

v = voxi + voyi +ay*T

ay = -32 j

v = (204*cos54)i + (204*sin54)j - 32*11.56 j

v = 120i + 165j - 369.92j

v = 120i - 204.92 j

v = sqrt(120^2+204.92^2) = 237.5 feet/s

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