A mortar crew is positioned near the top of a steep hill. Enemy forces are charg

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 = 56.0° (as shown), the crew fires the shell at a muzzle velocity of 137 feet per second. How far down the hill does the shell strike itf the hill subtends an angle = 32.0° from t horizontal? (Ignore air friction.) Vo 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

Explanation / Answer

The easiest path to solution is through the trajectory equation:
y = h + x·tan - g·x² / (2v²·cos²)
where y = height at given value of x; here y = -xtan
and h = initial height, let's say 0 ft
and x = range = ???
and = launch angle = 56º
and v = launch velocity = 137 ft/s

a)
Dropping units for ease,
-xtan32.0 = 0 + xtan56.0 - 32x² / (2(137)²cos²56)  
=> x = 773 ft horizontal range

d = x / cos = 773ft / cos32 = 911.5 ft
and this converts to d = 278 m

b)
flight time t = x / Vcos = 773ft / 137ft/s*cos56º = 10.1 s

c)
Vx = Vcos = 137ft/s * cos56º = 76.61 ft/s constant
Vy = Vsin + at = 137ft/s*sin56º - 32ft/s²*10.1s = -209.6 ft/s
speed = (Vx² + Vy²)
= sqrt(76.61^2 + (-209.6)^2)
= 223.2 ft/s = 68 m/s

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