A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m ,

Question

A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one with mass mA = 1.25 kg and the other with mass mB = 0.300 kg . In the explosion, 880 J of chemical energy is converted to kinetic energy of the two fragments.

What is the speed of each fragment just after the explosion? Va? and Vb?

It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Explanation / Answer

Here ,

at the maximum height , as the velocity of fragments will be zero before explosion

after the explosion ,

let the speeds is Va and Vb

0.50 * 1.25 * Va^2 + 0.50 * 0.30 * Vb^2 = 880

1.25 * Va - 0.30 * Vb = 0

solving for Va and Vb

Va = 16.5 m/s

Vb = 68.8 m/s

---------------------------------

NOw, time of flight , t = sqrt(2 * h/g)

t = sqrt(2 * 90/9.8)

t = 4.29 s

Now, for the distance between the points on ground

distance between the points on ground = (Va + Vb) * time

distance between the points on ground = (16.5 + 68.8) * 4.29

distance between the points on ground = 365.6 m

the distance between the points on ground is 365.5 m

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