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 = 64.0o (as shown), the crew fires the shell at a muzzle velocity of 2.00 × 102 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 33.0o from the horizontal? (Ignore air friction.) (b) how long will the mortar shell remain in the air. (c) how fast will the shell be traveling when it hits the ground?

Explanation / Answer

(a)

From 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 = 0 ft
and x = range = ???
and = launch angle = 64.0º
and v = launch velocity = 200 ft/s

Dropping units for ease,
-xtan(33.0) = 0 + xtan64.0 - 32x² / (2(200)²cos²64.0)
0 = 2.70x - 2.08*10-3*x²

x = 2.70 / 2.08*10-3 = 1297.93 ft horizontal range

d = x / cos = 1297.93 ft / cos33.0 = 1547.61 ft


(b)

flight time t = x / Vcos = 1297.93 ft/ 200 ft/s*cos64.0º = 14.80 s

(c)

Vx = Vcos = 200ft/s * cos64.0º = 87.67 ft/s
Vy = Vsin + at = 200ft/s*sin64.0º - 32ft/s²*14.80s = -293.84 ft/s
speed = (Vx² + Vy²) = 306.64 ft/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.