DA Sapling Learning maermen learning A mortar crew is positioned near the top of

Question

DA Sapling Learning maermen learning A mortar crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and necessary for the crew to spring into action. Angling the mortar at an angle of 6 61.0 (as shown), the crew fires the shell at a muzzle velocity of 264 feet per second. How far down the hill does the shell strike if the hill subtends an angle 36.00 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

y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-36º)
h = 0
x = ?
= 61º
v = 264 ft/s
x * sin(-36) = 0 + xtan61 - 32.2x² / (2*264²*cos²61)
-0.587x = 1.8040x - 0.0098x²
0 = 2.391x - 0.0098x²
x = 0 ft, 243.979 ft

So what does "down the hill" mean? Along the slope, it's 243.979ft/cos(-36º) = 1572 ft

y = 1572 * sin(-39) = -923.99 ft

time at/above launch height = 2·Vo·sin/g = 2 * 264ft/s * sin61 / 32.2 ft/s² = 14.3 s
initial vertical velocity Vy = 264ft/s * sin61º = 230.89 ft/s
so upon returning to launch height, Vv = -230.89 and time to reach the ground is
-923 ft = -230.89 * t - ½ * 32.2ft/s² * t²
0 = 923 - 230.89t - 16.1t²
quadratic; solutions at
t = -17.6444
3.2694
To the total time of flight is 14.3s + 3.2s = 17.5 s

at impact, Vy = Vvo- at = -230.89ft/s - 32.2ft/s² * 3.26s =-230.89ft/s -104.972 ft/s =-335.86ft/s
Vx = 264ft/s * cos61º = 127.98 ft/s
V = ((Vx)² + (Vy)²) = 359.417 ft/s = 109.55m/s

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