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 ? = 59.0o (as shown), the crew fires the shell at a muzzle velocity of 137 feet per second. How far down the hill does the shell strike if the hill subtends an angle ? = 34.0o 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

Distance along the hill = 266.16 m

Time taken = 220.66/21.507 = 10.2584 seconds...

Vertical component of velocity = 35.79 -10.26*9.8 = 64.76 m/s

Total vel = sqrt(64.76^2 + 21.51^2) = 68.24 m/s

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