Question 11 of 13 Sapling Learning Map do A mortar crew is positioned near the t

Question

Question 11 of 13 Sapling Learning Map do 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 68.00 (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 40.00 from the horizontal? (Ignore air Number 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 ) Check Answer Next Exit Hint

Explanation / Answer

along horizontal displacement = x = d*cosphi = d*cos33

initial speed = vox = vo*costheta

x = vox*T

T = x/vox = (d*cosphi)/(vo*costheta)

along vertical

displacement y = -d*sinphi

initial velocity = voy = vo*sintheta

y = voy *T + 0.5*ay*T^2

-d*sinphi = (vo*sintheta*d*cosphi)/(vo*costheta) - 0.5*g*d^2*(cosphi)^2/(vo^2*(costheta)^2)

-d*sinphi = (tantheta*d*cosphi)) - 0.5*g*d^2*(cosphi)^2/(vo^2*(costheta)^2)

-d*sin40 = (d*tan68*cos40)-(0.5*32*d^2*(cos40)^2)/(225^2*(cos68)^2)

d = 1921 feet

d = 1921*0.3048 m

d = 585.52 m           <<<<------------ANSWER

+++++++++++++++

T = (d*cos39)/(vo*cos67)

T = (1921*cos40)/(225*cos68)

T = 17.4 s       <<<<------------ANSWER

++++++++++++

along horizontal

vx = vox + ax*T = vox = 225*cos68 = 82.3 ft/s

along vertical

vy = voy +ay*T

ay = -32 j

v = ( (225*sin68) - 32*17.4

v = -348.2 ft/s

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

v = sqrt(84.3^2+348.2^2) = 358.26 feet/s

v = 358.26*0.3048 m/s

v = 109.2 m/s     <<<<------------ANSWER

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