don\'t copy someone elses answer i need a good explanation and steps A mortar* c

Question

don't copy someone elses answer i need a good explanation and steps

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 = 49.0degree (as shown), the crew fires the shell at a muzzle velocity of 256 feet per second. How far down the hill does the shell strike if the hill subtends an angle Greek phi = 39.0degree 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? Incorrect.

Explanation / Answer

1)

Vo=256 ft/s = 78.0288 m/s

consider x axis along length d and y perpendicular

acc. alog d = 9.81*sin(39 degree)=ax=6.173633 m/s^2

acc. perpendicular to d = ay= -9.81*cos(39 degree)=-7.6238 m/s^2

so,

Sy=0=Uy*t+1/2*ay*t^2

Uy*t=-1/2*ay*t^2

t=-Uy/(1/2*ay) = 78.0288*sin((49+39) degree)/(0.5*7.6238) = 20.4573223231 sec

so,

d=Sx=Ux*t+1/2*ax*t^2 = 78.0288*20.4573223231*cos((49+39) degree)+0.5*6.173633*20.4573223231^2 =1347.55 m

d = 1347.55 m

2)

t=20.4573223231 sec = 20.46 sec

3)

Vx=Ux+axt = 78.0288*cos((49+39) degree)+6.173633*20.46 = 129.035697028 m/s

Vy = Uy+ay*t = 78.0288*sin((49+39) degree)-7.6238*20.46 = -78.0016810367 m/s

|V|=sqrt(Vx^2+Vy^2) = sqrt(129.035697028^2+(-78.0016810367)^2) = 150.78 m/s

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