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 degrees, the crew fires the shell at a muzzle velocity of 127 feet per second. How far down the hill does the shell strike if the hill subtends an angle phi= 33.0 degrees 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? For the first question, the answer must be in meters. For the second, the answer must be in seconds. For the third, the answer must be in meters per second. A person already tried to answer this but the answers were incorrect. PLEASE HELP!

Explanation / Answer

Trajectory eqn: y = h + x·tanT - g·x² / (2v²·cos²T) y = x * sin(-33º) h = 0 x = ? T = 62º v = 127 ft/s x * sin(-33) = 0 + xtan62 - 32.2x² / (2*127²*cos²62) -0.545x = 1.88x - 0.00453x² 0 = 2.425x - 0.00453x² x = 0 ft, 535 ft So what does "down the hill" mean? Along the slope, it's 535ft/cos(-33º) = 638 ft EDIT: made an error here calculating y. Now fixed. y = 535 * sin(-33) = -291.4 ft time at/above launch height = 2·Vo·sinT/g = 2 * 127ft/s * sin62 / 32.2 ft/s² = 6.96 s initial vertical velocity Vv = 127ft/s * sin62º = 112.13 ft/s so upon returning to launch height, Vv = -112.13 and time to reach the ground is -291.4 ft = -112.13 * t - ½ * 32.2ft/s² * t² 0 = 291.4 - 112.13t - 16.1t² quadratic; solutions at t = 2.01 s, -8.98 s To the total time of flight is 6.96s + 2.01s = 8.97 s --> Answer at impact, Vv = Vvo * at = -112.13ft/s - 32.2ft/s² * 2.01s = -117.85 ft/s Vx = 127ft/s * cos62º = 59.62 ft/s V = v((Vx)² + (Vy)²) =132 ft/s--> Answer

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