I have tried this problem over and over, but no success.. Please show work. A mo
ID: 2232219 • Letter: I
Question
I have tried this problem over and over, but no success.. Please show work.
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 = 61.0 degree (as shown), the crew fires the shell at a muzzle velocity of 245 feet per second. How far down the hill does the shell strike if the hill subtends an angle phi = 38.0 degree 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? The mortar is like a small cannon that launches shells at steep angles.Explanation / Answer
Trajectory eqn: y = h + x·tanT - g·x² / (2v²·cos²T) y = x * sin(-38º) h = 0 x = ? T = 61º v = 245 ft/s x * sin(-38) = 0 + xtan61 - 32.2x² / (2*245²*cos²61) -0.615 x = 1.80x - 0.001141x² 0 = 2.415x - 0.001141x² x = 0 ft, 2116.56 ft So what does "down the hill" mean? Along the slope, it's 2116.56ft/cos(-38º) = 2686 ft (if you want in m , it is 818.7 m) EDIT: made an error here calculating y. Now fixed. y = 2686 * sin(-38) = -1653.63 ft (if you want in m , it is 504.025 m) time at/above launch height = 2·Vo·sinT/g = 2 * 245ft/s * sin61 / 32.2 ft/s² = 13.3 s initial vertical velocity Vv = 245ft/s * sin61º = 214.28 ft/s so upon returning to launch height, Vv = -214.28 and time to reach the ground is -1653.63 ft = -214.28 * t - ½ * 32.2ft/s² * t² 0 = 1653.63 - 214.28t - 16.1t² quadratic; solutions at t = 5.47 s, -18.77 s To the total time of flight is 13.3s + 5.47s = 18.77 s at impact, Vv = Vvo * at = -214.28ft/s - 32.2ft/s² * 5.47s = -390.41 ft/s Vx = 245ft/s * cos61º = 118.77 ft/s V = v((Vx)² + (Vy)²) = 408.07 ft/s (If you want answer in m/s, it is 124.38 m/s)