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

Question

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 -53.0° (as shown), the crew fires the shell at a muzzle velocity of 225 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 32.0° from the horizontal? (Ignore air Number 2614.5 IIn How long will the mortar shell remain in the air? Number 16.4 How fast will the shell be traveling when it hits the ground? Number 373.5 m/s ncorrect.

Explanation / Answer

use the trajectory equation:

y = h + x·tan - g·x² / (2v²·cos²)

where y = height at x-value of interest = -x*tan32.0º

and h = initial height = 0 m

and x = range of interest = ???

and = launch angle = 53.0º

and v = launch velocity = 225 ft/s

Dropping units for ease,

-x*tan32º = 0 + xtan53º - 32x² / (2*225²*cos²53º)

-x*tan32º = 0 + xtan53º - x² / 1146

-0.6249x = 1.327x - x²/1146

0 = 1.9519x - x²/1146 = x(1.9519 - x/1146)

has a trivial solution at x = 0

and another at x = 2237 ft

and so

y = vertical drop = -2237ft * tan32.0º = -1398 ft

and the distance downslope is

d = (x² + y²) = 2638 ft

= 2638 x 0.3048 m = 804m down the hill

time of flight

t = d / vcos57.0º = 2638ft / (225ft/s*cos53.0º) = 19.5 s

speed at impact

V = (v² + 2gh) = ((225ft/s)² + 2*32ft/s²*1398ft) = 374.3 ft/s

= 374.3 x 0.3048 = 114 m/s

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