Could you please answer the three questions above? Thank you s php?id 3390899 s

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Could you please answer the three questions above? Thank you s php?id 3390899 s Google Drive G Calculus 8th Edition c Stewart caladus Early O Mylab&Masteringl-; C 3rd ledbook Maies Print caloulator Edwton Periodic Table estion 11 of 13 In 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 0 55.00 (as shown), the crew fires the shell at a muzzle velocity of 129 feet per second. How far down the hill does the shell strike the hill subtends an angle 34.00 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? m/s Previous Check Answer Next HE

Explanation / Answer

Trajectory eqn:
y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-34º)
h = 0
x = ?
= 55º
v = 129 ft/s
x * sin(-34) = 0 + xtan55 - 32.2x² / (2*129²*cos²55)
-0.559x = 1.42x - 0.00294x²
0 = 1.422x - 0.00294x²
x = 0 ft, 483.67 ft

So what does "down the hill" mean? Along the slope, it's 483.67ft/cos(-34º) = 583.41 ft

time at/above launch height = 2·Vo·sin/g = 2 * 129ft/s * sin55 / 32.2 ft/s² = 6.56 s
initial vertical velocity Vv = 129ft/s * sin55º = 105.67 ft/s
so upon returning to launch height, Vv = -105.67 and time to reach the ground is
-583.41 ft = -105.67 * t - ½ * 32.2ft/s² * t²
0 = 583.41 - 105.67t - 16.1t²
quadratic; solutions at
t = 3.57 s
To the total time of flight is 6.56s + 3.57s = 10.13 s

at impact, Vv = Vvo * at = -105.67ft/s - 32.2ft/s² * 3.57s = -220.624 ft/s
Vx = 129ft/s * cos55º = 73.99 ft/s
V = ((Vx)² + (Vy)²) = 232.700 ft/s

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