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 = 49.0o (as shown), 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 = 33.0o 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 equation: y = h + x·tan - g·x² / (2v²·cos²)

y = x * sin(-33º)
h = 0
x = ?
= 49º
v = 127 ft/s
x * sin(-33) = 0 + xtan49 - 32.2x² / (2*127²*cos²49)
-0.544x = 1.15x - 0.0020x²
0 = 1.69x - 0.0020x²
x = 0 ft, 1274 ft

So what does "down the hill" mean? Along the slope, it's 1274ft/cos(-33º) = 1520 ft

EDIT: made an error here calculating y. Now fixed.
y = 1520 * sin(-33) = -827 ft

time at/above launch height = 2·Vo·sin/g = 2 * 127ft/s * sin49 / 32.2 ft/s² = 5.9 s
initial vertical velocity Vv = 127ft/s * sin49º = 95.8 ft/s
so upon returning to launch height, Vv = -95.8 and time to reach the ground is
-827 ft = -95.8 * t - ½ * 32.2ft/s² * t²
0 = 827 - 95.8t - 16.1t²
quadratic; solutions at
t = 5.34 s, -11.2 s
To the total time of flight is 5.9s + 5.3s = 11.2 s

at impact, Vv = Vvo * at = = -95.8ft/s - 32.2ft/s² * 5.3s = -266 ft/s
Vx = 127ft/s * cos49º = 83.3 ft/s
V = ((Vx)² + (Vy)²) = 252 ft/s

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