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

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 - 0.000599x²
0 = 1.79x - 0.000599x²
x = 0 ft, 1986ft

So what does "down the hill" mean? Along the slope, it's 1986ft/cos(-37º) = 2486.74 ft =757.95 meters
here calculating y. Now fixed.
y = 2486.74 * sin(-37) = -1496.55 ft =-456.14 meters

time at/above launch height = 2·Vo·sin/g = 2 * 255ft/s * sin50 / 32.2 ft/s² = 12.13 s
initial vertical velocity Vv = 255ft/s * sin50º = 195.34 ft/s
so upon returning to launch height, Vv = -195.34 and time to reach the ground is
-1496.55 ft = -195.34 * t - ½ * 32.2ft/s² * t²
0 = 1496.55 - 195.34t - 16.1t²
quadratic; solutions at
t = 5.32 s, -17.45 s
To the total time of flight is 12.13s + 5.32s = 17.45s

at impact, Vv = Vvo * at = -195.34ft/s - 32.2ft/s² * 5.32s = -366.64 ft/s
Vx = 255ft/s * cos50º = 163.91 ft/s
V = ((Vx)² + (Vy)²) = 401.33 ft/s = 122.32 m/s

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