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 e 58.0 (as shown), the crew fires the shell at a muzzle velocity of 217 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 35.00 from the horizontal? (Ignore air friction.) Number How long will the mortar shell remain in the air? Number How fast will the shell be traveling when it hits the ground? Number m/s

Explanation / Answer

Velocity=66.14 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= -xtan 35 = -0.7x ---(i)

the horizontal component of velocity= 66.14* cos 58 =35 m/s.

the vertical component of velocity initially=66.14 *sin58= 56.09 m/s.

The equation of trajectory of a projectile is given by

y= x tan58 - 9.8 x2Sec266/(2*v2)

y= 1.6 x- 3.99*10^-3*x2

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

-0.7 x =1.6x - 3.99*10^-3*x2

Therefore, x=401 m.

The distance along the hill= 401 Sec 35=489.5 m.

Time taken to reach the final point= 489.5/35=13.98 seconds.

Vertical compoenent of velocity then= 56.09 -13.98*9.8

= -81.05m/s

therefore speed= sqrt( 81.052+352)

=88.28 m/s

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