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 = 63.0o (as shown), the crew fires the shell at a muzzle velocity of 193 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 31.0o 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 eqn:
y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-31º)
h = 0
x = ?
= 63º
v = 193 ft/s
x * sin(-31) = 0 + xtan63 - 32.2x² / (2*193²*cos²63)
x = 0 ft, 1272.2 ft

Along the slope, it's 1272.2ft/cos(-31º) = 1484.1 ft

y = 1484.1 * sin(-31) = -764.3 ft

time at/above launch height = 2·Vo·sin/g = 2 * 193ft/s * sin63 / 32.2 ft/s² = 10.6 s
initial vertical velocity Vv = 193ft/s * sin63º = 171.9 ft/s
so upon returning to launch height, Vv = -171.9 and time to reach the ground is
-764.3 ft = -171.9 * t - ½ * 32.2ft/s² * t²
0 = 764.3 - 171.9t - 32t²
quadratic; solutions at
t = 2.89 s, -8.26 s
To the total time of flight is 9.8s + 2.89s = 12.6 s

at impact, Vv = Vvo * at = -171.9ft/s - 32.2ft/s² * 2.89s = 15996.6 ft/s
Vx = 193ft/s * cos63º = 87.6 ft/s
V = ((Vx)² + (Vy)²) = 342 ft/s

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