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?

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.