(adsbygoogle = window.adsbygoogle || []).push({}); 1. A 7.1 kg firework is launched straight up and at its maximum height 29 m it explodes into three parts. Part A(1 kg) moves straight down and lands 0.28 seconds after the explosion. Part B(2.5 kg) moves horizontally to the right and lands 12.5 meters from Part A. Part C moves to the left at some angle. How far from Part A does Part C land(no direction needed)?

2. x-x_0=v_0*\cos{theta}*t, y-y_0=v_0*\sin{theta}*t+0.5*a*t^{2}

momentum is conserved

3. I first figured out the initial velocity of A, by using the second equation above, namely -29=v_0*sin(-90)*0.28+0.5*0.28^2, and solved for v_0 of a

Then I solved v0 of B, from the first equation above, namely 12.5=v_0*cos(0)*0.28

Then I could solve for the initial velocity of part C and the direction, by using vectors on the momentum.

Then I can figure out the distance from C to A by the two formulas above, but I keep getting the wrong answer.

Thanks

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# Homework Help: Determining Distance

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