A firework shoots off at. an angle of theta_0 = 53 degree at a speed pf v_0 = 20

Question

A firework shoots off at. an angle of theta_0 = 53 degree at a speed pf v_0 = 20 m/s. At the apex of the firework's flight it explodes into two pieces, say one if blue and the other is green. The green firework leaves the explosion at an angle of theta_G = 37 degree with a speed of 15 m/s. The total mass of the firework had a mass of 1 kg, the green piece had a mass of 0.7 kg and the blue piece had a mass of 0.3 kg. The explosion was generated by a negligible amount of black powder. Calculate the height at which the firework explodes. Calculate the vertical and horizontal velocity of the firework at the apex of its flight. Calculate the horizontal and vertical velocity of the blue piece after the explosion. Calculate the horizontal distance where the green piece lands on the ground relative to where the firework we launched Calculate the horizontal distance where the blue piece lands on the ground relative to where the firework launched. How far apart did the green and blue pieces land from one another?

Explanation / Answer

e)

mb = mass of blue peice = 0.3 kg

mg = mass of green piece = 0.7 kg

M = total mass before explosion = 1 kg

Using conservation of momentum along X-direction :

M Vo Cos53 = mb Vbx + mg (15 Cos37)

1 (20) Cos53 = 0.3 Vbx + (0.7) (15 Cos37)

Vbx = 12.2 m/s

Using conservation of momentum along X-direction :

M (0) = mb Vby + mg (15 Sin37)

0 = 0.3 Vby + (0.7) (15 Sin37)

Vby = - 21.1 m/s

horizontal distance of the apex

X = Vo2 Sin2(53) /2g = 202 Sin106 /(2 x 9.8) = 19.62 m

vertical height at which explosion takes place

Y = Vo2 Sin2(53) /2g = 202 Sin253 /(2 x 9.8) = 13.02 m

consider the motion of blue ball in vertical direction

Y = Vby t + (0.5) g t2

13.02 = 21.1 t + (0.5) (9.8) t2

t = 0.55

distanceof landing point of blue piece from initial launching point

Dx = X + Vbx t = 19.62 + 12.2 (0.55) = 26.33 m

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