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 ? = 65.0o (as shown), the crew fires the shell at a muzzle velocity of 246 feet per second. How far down the hill does the shell strike if the hill subtends an angle ? = 32.0o 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

y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-32º)
h = 0
x = ?
= 65º
v = 246 ft/s
x * sin(-32) = 0 + xtan65 - 32.2x² / (2*246²*cos²65)
-0.53x = 2.145x - 0.0015x²
0 = 2.675x - 0.0015x²
x = 0 ft, 1783 ft

So what does "down the hill" mean? Along the slope, it's 1783ft/cos(-32º) = 2102 ft
y = 2102 * sin(-32) = -1114 ft

time at/above launch height = 2·Vo·sin/g = 2 * 246ft/s * sin65 / 32.2 ft/s² = 13.8 s
initial vertical velocity Vv = 246ft/s * sin65º = 222.95 ft/s
so upon returning to launch height, Vv = -222.95 and time to reach the ground is
-1114 ft = -222.95 * t - ½ * 32.2ft/s² * t²
0 = 1114 - 222.95t - 16.1t²
quadratic; solutions at
t = 3.9 s, -17.8 s
To the total time of flight is 13.8s + 3.9s = 17.7 s

at impact, Vv = Vvo * at = -222.95ft/s - 32.2ft/s² * 3.9s = - 348.5 ft/s
Vx = 246ft/s * cos65º = 103.96 ft/s
V = ((Vx)² + (Vy)²)

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