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 = 57.0° (as shown), the crew fires the shell at a muzzle velocity of 262 feet per second. How far down the hill does the shell strike i the hill subtends an angle = 40.00 from the horizontal? (Ignore air friction.) Number Im 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

perpendicular to the incline,

y = v0y t + ay t^2 / 2

0 = (262 sin(57 + 40))t - (32.17 cos40) t^2 /2

t = 21.1 sec ......Ans

x = (262 cos(57 + 40))(21.1) + (32.174 cos40)(21.1^2 / 2)

x = 6490 ft = 2000 m .......Ans

h = x sin40 = 1284 m

v0 = 262 x 0.3048 =79.9 m/s

v = sqrt[ 79.9^2 + (2 x 9.8 x 1284)]

v =177.6 m/s ........Ans

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