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saplinglearning.com SEN: HW.. www.saplinglearning.com/ibis/css/Accessibility.php? has just finished climbing a sheer cliff ab y- PHYS 2425-Spring18-SEN Activities and Due Dates > HW: 2D &3D Motion 2/12/2018 11:59 PM O 48.6/100 Gradeb Print Calculator Periodc Table Question 6 of 8 Sapling Learning Map ofh 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 = 65.00 (as shown), the crew fires the shell at a muzzle velocity of 261 feet second. How far down the hill does the shell strike the hill subtends an angle 34.0° from the horizontal? (Ignore air friction.) Number In 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 O Previous Give Up & View Sousona Check Answer ONext t Hint Exit

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*tan34.0º
and h = initial height = 0 m
and x = range of interest = ???
and = launch angle = 65.0º
and v = launch velocity = 261 ft/s

Dropping units for ease,
-x*tan34º = 0 + xtan65º - 32x² / (2*261²*cos²65º)
-x*tan34º = 0 + xtan65º - x² / 760.427
-0.6745x = 2.1445x - x²/760.427
0 = 2.819x - x²/760.427= x(2.819 - x/760.427)
has a trivial solution at x = 0
and another at x = 2143.6437 ft
and so
y = vertical drop = -2143.6437 ft * tan34.0º = -1445.906 ft
and the distance downslope is
d = (x² + y²) = 2585.7 ft = 788.121 m down the hill (Answer)

time of flight
t = x / vcos65.0º = 2143.6437 ft / (261 ft/s*cos65.0º) = 19.43 s (Answer)

speed at impact
V = (v² + 2gh) = ((261ft/s)² + 2*32ft/s²* 1445.906 ft) = 400.82 ft/s = 122.17 m/s (Answer)

Thanks!!!

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