Map Sapling Learning A mortar* crew is positioned near the top of a steep hill.

Question

Map Sapling Learning 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 = 58.00 (as shown), the crew fires the shell at a muzzle velocity of 217 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 35.00 from the horizontal? (Ignore air Number 498.531 Im How long will the mortar shell remain in the air? Number 13.967 How fast will the shell be traveling when it hits the ground? Number 88.063 m/ s Incorrect.

Explanation / Answer

Solution:

Let us go to the basics first.

Use the trajectory equation:
y = h + x·tan - g·x² / (2v²·cos²)
where y = height at x-value of interest = -x*tan35.0º
and h = initial height = 0 m
and x = range of interest = ???
and = launch angle = 58.0º
and v = launch velocity = 217 ft/s

Dropping units for ease,
-x*tan35º = 0 + xtan58º - 32x² / (2*217²*cos²58º)
-x*tan35º = 0 + xtan58º - x² / 826.454
-0.7x = 1.6x - x²/826.454
0 = 2.3x - x²/826.454 = x(2.3 - x/826.454)
has a trivial solution at x = 0
and another at x = 1900.8442 ft
and so
y = vertical drop = -1900.8442 ft * tan35.0º = -1330.985 ft
and the distance downslope is
d = (x² + y²) = 2320.50 ft = 707.288 m down the hill (Answer)

time of flight
t = x / vcos58.0º = 1900.8442 ft / (217 ft/s*cos58.0º) = 16.53 s (Answer)

speed at impact
V = (v² + 2gh) = ((217ft/s)² + 2*32ft/s²* 1330.985 ft) = 363.7 ft/s = 110.85 m/s (Answer)

Thanks!!!

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