mortar* crew is positioned near the top of a steep hill. Enemy forces are chargi

Question

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 = 52.0digree (as shown), the crew fires the shell at a muzzle velocity of 168 feet per second. How far down the hill does the shell strike if the hill subtends an angle =31.0digree 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

y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-39º)
h = 0
x = ?
= 58º
v = 184 ft/s
x * sin(-39) = 0 + xtan58 - 32.2x² / (2*184²*cos²58)
-0.629x = 1.66x - 0.0018x²
0 = 2.29x - 0.0018x²
x = 0 ft, 1274 ft

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

y = 1640 * sin(-39) = -1302 ft

time at/above launch height = 2·Vo·sin/g = 2 * 184ft/s * sin58/ 32.2 ft/s² = 9.8 s
initial vertical velocity Vv = 184ft/s * sin58º = 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 * cos58º = 94.8 ft/s
V = ((Vx)² + (Vy)²) = 342 ft/s

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