Map Sapling Learning A mortar crew is positioned near the top of a steep hill. E

Question

Map Sapling Learning 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 61.0° (as shown), the crew fires the shell at a muzzle velocity of 262 feet per second. How far down the hill does the shell strike the hill subtends an angle = 37.00 from the horizontal? (Ignore air friction.) Number How long will the mortar shell remain in the air? Number How fast will the shell be traveling when it hits the ground? Number m/ s ) Previous Give Up & View Solution Check Answer Next Exit- Hint

Explanation / Answer

Trajectory eqn:
y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-37º)
h = 0
x = ?
= 61º
v = 262 ft/s
x * sin(-37) = 0 + xtan61 - 32.2x² / (2*262²*cos²61)
-0.60x = 1.8x - 0.000997x²
0 = 2.4x - 0.000997x²
x = 0 ft, 2407.22 ft

So what does "down the hill" mean? Along the slope, it's 2407.22 ft/cos(-37º) = 3014.16 ft


y = 3014.16 * sin(-37) = -1813 ft

time at/above launch height = 2·Vo·sin/g = 2 * 262ft/s * sin61 / 32.2 ft/s² = 14.23 s
initial vertical velocity Vv = 262ft/s * sin61º = 229.15 ft/s
so upon returning to launch height, Vv = -229.15 and time to reach the ground is
-1813 ft = -229.15 * t - ½ * 32.2ft/s² * t²
0 = 1813 - 229.15 - 16.1t²
quadratic; solutions at
t = 5.66 s, -19.89 s
To the total time of flight is 14.23s + 5.66s = 19.89 s

at impact, Vv = Vvo * at = -229.15ft/s - 32.2ft/s² * 5.66s = -411.4 ft/s
Vx = 262ft/s * cos61º = 127 ft/s
V = ((Vx)² + (Vy)²) = 430.56 ft/s

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