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=64.0, the crew fires the shell at a muzzle velocity of 228 feet per second. How far down the hill does the shell strike if the hill subtends an angle phi=38.0 from the horizontal? (ignore air friction.) How long will the mortar remain in the air? How fast will the shell be traveling when it hits the ground?

Explanation / Answer

we use the kinematic equations,

(a) The equation of the hill is y= -xtan 32= -0.625x ---(i)

v= 253 ft/s=77.11 m/s

the horizontal component of velocity= 77.11* cos56=43.11 m/s.

the vertical component of velocity initially=77.11 *sin 56= 63.9 m/s.

The equation of trajectory of a projectile is given by

y= x tan 56- 9.8 x2Sec256(2*v2)

y= 1.483 x- 2.635*10^-3*x2

Substituting, (i) in this equation gives the x coordinate of the landing point.

-0.625 x =1.483x - 2.635*10^-3*x2

Therefore, x=800 m.

The distance along the hill= 800Sec 32=943 m.

(b) Time taken to reach the final point= 800/43.11=18.55 seconds.

(c) Vertical compoenent of velocity then Vy= Voy-gt= 63.9 -18.55*9.8

= -117.89m/s

final speed= sqrt( 117.89+43.112)

Vf=125.5 m/s

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