A 2.97-kg projectile is fired with an initial speed of 133 m/s at an angle of 28

Question

A 2.97-kg projectile is fired with an initial speed of 133 m/s at an angle of 28° with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.97 kg and 2 kg. At 3.87 s after the explosion, the 2-kg fragment lands on the ground directly below the point of explosion.
(a) Determine the velocity of the 0.97-kg fragment immediately after the explosion.
1 = m/s + m/s

(b) Find the distance between the point of firing and the point at which the 0.97-kg fragment strikes the ground.
km

(c) Determine the energy released in the explosion.
kJ

Explanation / Answer

time required to reach at the highest point= 133sin(28)/g=6.36 sec velocity of projectile at highest point= 133 cos(28)=117.4 then apply momentum conservation, 2.97*117.4=.97*0+2*v or,v=174.3 m/s

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