A mortar* crew is positioned near the top of a sleep hill. Enemy forces are char

Question

A mortar* crew is positioned near the top of a sleep hill. Enemy forces are charging up the hill and il is necessary for the crew to spring into action. Angling the mortar at an angle of 8 = 65. 0 degree (as shown), the crew fires the shell at a muzzle velocity of 232 feet per second. How far down the hill does the shell strike if the hill subtends an angle phi = 39. 0 degree 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? * The morlar is like a small cannon that launches shells al steep angles.

Explanation / Answer

Trajectory eqn:
y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-39º)
h = 0
x = ?
= 65º
v = 232 ft/s
x * sin(-39) = 0 + xtan65 - 32.2x² / (2*232²*cos²65)
-0.629x = 2.14x - 0.00167x²
0 = 2.769x - 0.00167x²
x = 0 ft, 1658 ft

So what does "down the hill" mean? Along the slope, it's 1658ft/cos(-39º) = 2133 ft

y = 2133* sin(-39) = -1342.62 ft

time at/above launch height = 2·Vo·sin/g
= 2 * 232ft/s * sin65 / 32.2 ft/s²
= 13.06 s
initial vertical velocity Vv = 232ft/s * sin65º = 256 ft/s
so upon returning to launch height, Vv = -256 and time to reach the ground is
-1342.62 ft = -256 * t - ½ * 32.2ft/s² * t²
0 = 1342.62 - 256t - 16.1t²
quadratic; solutions at
t = 4.16 s, -20.06 s
To the total time of flight is 13.06s +4.16s = 17.22 s

at impact, Vv = Vvo * at = -256ft/s - 32.2ft/s² * 4.16s = -390 ft/s
Vx = 232ft/s * cos65º = 98.05 ft/s
V = ((Vx)² + (Vy)²) = 402 ft/s

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