A 12.0-kg shell is launched at an angle of 55.0 ∘ above the horizontal with an initial speed of 150 m/s. when it is at its highest point, the shell exploded into two fragments, one three times heavier than the other. the two fragments reach the ground at the same time. assume that air resistance can be ignored. if the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land and how much energy was released in the explosion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

As we know that projectile is launched at an angle of 55 degree with the horizontal so at highest point of its path the velocity of projectile is only along horizontal direction

so we will have

[tex]v_x=vcos55[/tex]

[tex]v_x=150cos55=86.03 \ \dfrac{m}{s}[/tex]

Now it explodes into two equal parts so by momentum conservation we will have

[tex]mv_x=\dfrac{m}{2}v[/tex]

[tex]v=2v_x[/tex]

[tex]v=2\times 86.03=172.03\ \frac{m}{s}[/tex]

[tex]T=\dfrac{2vSin\theta}{g}[/tex]

now total distance moved by it is given as

[tex]R=\dfrac{vsin55}{g}(vcos55)+\dfrac{vsin55}{g}(2vcos55)[/tex]

[tex]R=\dfrac{172.07^2(0.5735)(0.8191)}{9.81}+\dfrac{2(172.03)^2)(0.5735)(0.8191)}{9.81}[/tex]

[tex]R= 4248 \ m[/tex][tex]R=4248\ m[/tex]

(b) Energy was released in the explosion:

[tex]\Delta E=\dfrac{1}{2}\dfrac{m}{2}(2vcos55)^2-\dfrac{1}{2}m(vcos55)^2[/tex]

[tex]\Delta E=\dfrac{1}{4}mv^2-\dfrac{1}{8}mv^2[/tex]

[tex]\Delta E= \dfrac{1}{8}(12)(172.03^2)[/tex]

[tex]\Delta E=44391 \ J[/tex]

Hence The distance travel by the shell = 4248 n and The energy of the shell will be = 44391 J

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