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 232 feet per second. How far down the hill does the shell strike if the hill subtends an angle ? = 36.0o from the horizontal? (Ignore air friction.) (anwser in meters)How long will the mortar shell remain in the air ( anwesr in seconds)?How fast will the shell be traveling when it hits the ground(anwser in M/s)?

Explanation / Answer

How long will the mortar shell remain in the air?
t = 195.5985/4.9 = 39.918 sec

How fast will the shell be traveling when it hits the ground?
Vf = ?2gh+V^2

apply the formula u will get the answer

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