A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its t

Question

A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.83s after the explosion.

When does the second reach the ground?

Express your answer using two significant figures.

Now, I thought that because the masses are the same, the second mass would hit the ground in the same amount of time, and that seems to be the general consensus on here but it's incorrect :/ Any help would be greatly appreciated!

Explanation / Answer

When the fireworks reach the peak of its trajectory, the velocity of the firework is zero, hence:
v^2 = u^2 + 2as
0 = 1600 m^2 s^-2 - (2) (9.81) (s)
s =81.5 m (3s.f.)
During the explosion, the two parts gain equal vertical momentum in opposite directions.
Since the first piece lands in 2.83s,
s = ut + (0.5) at^2
81.5 = u (2.83) + (0.5) (9.81) (2.83^2)
u = 14.9 ms^-1 (3s.f.)
taking the origin as the point of explosion,
s = ut + 1/2 (at^2)
-81.5 = (14.9) (t) + 1/2 (9.81) (t^2)
t = 5.87s (3s.f.) or t = -2.83s (3s.f.) (rej)
ans: 5.87s after explosion, or 9.94s after the firework was launched

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