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 theta = 57.0 degree (as shown), the crew fires the shell at a muzzle velocity of 212 feet per second. How far down the hill does the shell strike if the hill subtends an angle Phi = 37.0 degree from the horizontal? (Ignore air friction.) How long will the mortar shell remain in the air? How fast will the shell be traveling when it hits the ground? The mortar is like a small cannon that launches shells at steep angles.

Explanation / Answer

given

vo = 212 ft/s

= 212*0.3048

= 64.6 m/s

let t is the time taken for mortar to hit the ground.

in x-direction

d*cos(37) = 64.6*cos(57)*t --------(1)

in y-direction

-d*sin(37) = 64.6*sin(57)*t - (1/2)*9.8*t^2 ----(2)

devide equation(2) with equation(1)

-tan(37) = (64.6*sin(57)*t - 4.9*t^2)/(64.6*cos(57*t))

-tan(37) = tan(57) - 4.9*t/(64.6*cos(57))

-(tan(37) + tan(57)) = -4.9*t/(64.6*cos(57))

t = (tan(37) + tan(57))*64.6*cos(57)/4.9

= 16.47 s

so, from equation 1

d*cos(37) = 64.6*cos(57)*16.47

d = 726 m <<<<<<<<<<<<<---------------------Answer

b) t = 16.47 s

c) vx = vox

= 64.6*cos(57)

= 35.2 m/s

vy = voy - g*t

= 64.6*sin(57) - 9.8*16.47

= -107.2 m/s

so,

v = sqrt(vx^2 + vy^2)

= sqrt(35.2^2 + 107.2^2)

= 113 m/s <<<<<<<<<<<<<---------------------Answer

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