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 = 60.0o (as shown), the crew fires the shell at a muzzle velocity of 238 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 40.0o 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

we use the kinematic equations,

(a) The equation of the hill is y= -xtan40= -0.84x ---(i)

v= 238 ft/s=72.54 m/s

the horizontal component of velocity= 72.53* cos60=36.3 m/s.

the vertical component of velocity initially=72.53 *sin 60= 62.8m/s.

The equation of trajectory of a projectile is given by

y= x tan 60- 9.8 x2Sec260/(2*v2)

y= 1.732 x- 3.73*10^-3*x2

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

-0.84x=1.732x - 3.73*10^-3*x2

Therefore, x=690m.

The distance along the hill= 690Sec40=900 m.

(b) Time taken to reach the final point= 690/36.3=19 seconds.

(c) Vertical compoenent of velocity then Vy= Voy-gt= 62.8 -19*9.8

= -123.4m/s

final speed= sqrt( 123.42+36.32)

Vf=128.6m/s

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