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 degree (as shown), the crew fires the shell at a muzzle velocity of 183 feet per second. How far down the hill does the shell strike if the hill subtends an angle phi = 31.0 degree 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

Velocity= 183 ft/s * 0.305 = 55.78 m/s.

Let the point from which the mortar is thrown be the origin of our coordinate system with the vertical direction the y

axis and downward being the negative y.

The equation of the hill is y= -x*tan(31) = -0.601*x ---(i)

the horizontal component of velocity= 55.78*cos(64) = 24.43 m/s.

the vertical component of velocity initially= 55.78 *sin(64) = 50.15 m/s.

The equation of trajectory of a projectile is given by

y= x*tan(64) - 9.8*x^2Sec^264)/(2*v^2)

y= 2.05*x- 10.13*10^-3*x^2

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

-0.601*x =2.05*x - 10.13*10^-3*x^2

Therefore, x = 261.7 m.

The distance along the hill= 261.7*Sec(31) = 305.4 m.

Time taken to reach the final point= 261.7/24.43 = 10.71 seconds.

Vertical compoenent of velocity then= 50.15 -10.71*9.8

= -54.808 m/s

therefore speed= sqrt( 54.808^2+24.43^2)

= 60 m/s

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