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 = 67.0o (as shown), the crew fires the shell at a muzzle velocity of 193 feet per second.

1). How far down the hill does the shell strike if the hill subtends an angle = 39.0o from the horizontal? (Ignore air friction.)

2). How long will the mortar shell remain in the air?

3). How fast will the shell be traveling when it hits the ground?

Explanation / Answer

along horizontal 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*sin39 = (d*tan67*cos39)-(0.5*32*d^2*(cos39)^2)/(193^2*(cos67)^2)

d = 1447.8 feet

d = 1447.8*0.3048 m

d = 441.3 m           <<<<------------ANSWER

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

T = (d*cos39)/(vo*cos67)

T = (1447.8*cos39)/(193*cos67)

T = 14.9 s       <<<<------------ANSWER

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

v = voxi + voyi +ay*T

ay = -32 j

v = (193*cos69)i + (193*sin67)j - 32*14.9 j

v = 69.1i + 177.6j - 476.8j

v = 69.1i - 299.2 j

v = sqrt(69.1^2 + 299.2^2) = 307 feet/s       <<<<------------ANSWER

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