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 = 68.0o (as shown), the crew fires the shell at a muzzle velocity of 219 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 34.0o 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?

Explanation / Answer

use the quation from projectile motion as

Y = tan(theta)*X - 0.5*9.8*X^2/V^2cos(theta)^2

Y = d*sin(68) = 0.927 d

X= d*cos(phi) = d*0.374

0.927d = 2.475*d*0.857 - d^2*0.0117

d = 102.50 m

--------------------------
1 ft = 0.09

along vertical Y = d*sinphi = 102.50 * 0.927 = 95.10 m

along horizontal X = d*cos(phi)= 102.5 * 0.374 = 38.35 m

but X = V*cos68*T

95.10 = (219 * 0.0929 * cos 68) T

T = 12.47 secs

Vfx =Vx = V *cos68   = 219* 0.0929 * cos 68 = 7.62 m/s

Vfy = Vy -gT = V*sin68-9.8*T = (219 *0.0929 * sin 68 - 9.8* 12.47) = -103.34 m/s

Vf = Vfx+Vfy

magnitude = sqrt(Vfx^2+Vfy^2) = sqrt(7.62^2 + 103.34^2)

Velocity = 103.62 m/s or 1115.39 ft/s

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