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= 62.0 degree (as shown), the crew fires the shell at a muzzle velocity of 208 feet per second. How far down the hill does the shell strike if the hill subtends an angle Phi = 40.0 degree from the horizontal? (Ignore air friction.) Number m How long will the mortar shell remain in the air? Number s How fast will the shell be traveling when it hits the ground? Number m/s

Explanation / Answer

x = d*cos40 = Vox*T

T = d*cos40/VO*cos62 = 0.0078d

y = d*sin40 =

y = Voy*T + 0.5*g*T^2

-d*sin40 = Vo*sin62*0.0078*d - 16*0.0078^2*d^2

-0.642d = 1.43d - 0.00097344d^2

d = 2128.533 ft = 649 m

T = 16.6 s

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