ortar crew is positioned near the top of a steep nil. Enemy forces are charging

Question

ortar crew is positioned near the top of a steep nil. 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 350.0 (as shown), the 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 37.0o from the horizontal? (Ignore air friction.) Number Tools x 10 shell remain in the air? How long Number How fast will the shell be traveling when it hits the ground? Number

Explanation / Answer

Trajectory eqn:
y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-37º)
h = 0
x = ?
= 50º
v = 255 ft/s
x * sin(-37) = 0 + xtan50 - 32.2x² / (2*255²*cos²50)
-0.601x = 1.19x - 0.0018x²
0 = 2.29x - 0.0005x²
x = 0 ft, 1274 ft

So what does "down the hill" mean? Along the slope, it's 1274ft/cos(-39º) = 1640 ft

EDIT: made an error here calculating y. Now fixed.
y = 1640 * sin(-39) = -1302 ft

time at/above launch height = 2·Vo·sin/g = 2 * 184ft/s * sin59 / 32.2 ft/s² = 9.8 s
initial vertical velocity Vv = 184ft/s * sin59º = 157.7 ft/s
so upon returning to launch height, Vv = -157.7 and time to reach the ground is
-1302 ft = -157.7 * t - ½ * 32.2ft/s² * t²
0 = 1302 - 157.7t - 16.1t²
quadratic; solutions at
t = 5.34 s, -15.1 s

To the total time of flight is 9.8s + 5.3s = 15.1 s

at impact, Vv = Vvo * at = -157.7ft/s - 32.2ft/s² * 5.3s = -328 ft/s
Vx = 184ft/s * cos59º = 94.8 ft/s
V = ((Vx)² + (Vy)²) = 342 ft/s

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