hi, I'd be really happy if someone could help me out with this question.?

Question

A flowerpot is dropped from a balcony of an apartment 28.5m above the ground. At a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0m above the ground. The initial velocity of the ball is 12m/s (down). Does the ball pass the flowerpot before striking the ground? If so, how far above the ground are the two objcts when the ball passes the flowerpot?

Answer 1

Let us suppose that they meet at a distance x from the ground and at a time t from the instant the pot is dropped.
[28.5 -x] = 0.5gt^2 for the pot. ---------------------1
[26 -x] = 12 [t -1] + 0.5 g [t-1] ^2 for the ball.
]
[26 -x] = 4.9 t^2 + 2.2 t - 7.1
From 1
[26 -x] = [28.5 -x] + 2.2 t - 7.1
2.2 t - 4.6 = 0
t = 2.09 s.
From [28.5 -x] = 0.5gt^2 = 21.40369
x = 7.09 m
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Both meet after time of 2.09 s at a distance
x = 7.09 m from the ground,
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Answer 2

Since the flower pot is simply dropped, its location relative to the second floor is
xp(t) = +2 - 4.9 * t^2
The ball's equation is
xb(t) = -12(t-1) -4.9*(t-1)^2,t>1
When (or if) they meet,
2-4.9t^2 = -12(t-1)-4.9(t-1)^2
Then 2 = -12t+12+9.8t-4.9
-5.1 = -2.2 t, so t= 2.3 sec appx.
Substitute this into the xp(t) eqtn