ortar* crew is positioned near the top of a steep nill. Enemy forces are chargin

Question

ortar* crew is positioned near the top of a steep nill. 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 = 50.0° (as shown), the Vo crew fires the shell at a muzzle velocity of 255 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 37.00 from the the hill subtends an angle d - 370° from the horizontal? (Ignore air friction.) Number Tools x 102 How long v shell remain in the air? Number How fast will the shell be traveling when it hits the ground? Number m/s

Explanation / Answer

y = h + x·tan - g·x² / (2v²·cos²)
where y = final height = 0 (somewhat arbitrarily)
and h = dsin = d*sin37º = 0.6d
and x = dcos = 0.8d
and = 50º
and v = 255 ft/s
0 = 0.6d + 0.8d*tan50 - 32ft/s² * (0.8d)² / (2(255)²cos²50)
0 = 2.24d - 0.003d² = d(2.24 - 0.003d)
first solution: d = 0 (trivial)
second solution: d = 2.24 / 0.003 = 755 ft answer

to find the time, divide the range (x) by the horizontal velocity:
t = 0.8 *d / (255ft/s * cos50) = ?s

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