- #1

MuchJokes

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**1. A flowerpot is dropped from the balcony of an apartment, 28.5 m 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, 26.0 m above the ground. The initial velocity of the ball is 12.0 m/s [down]. Does the ball pass the flowerpot before striking the ground? If so, how far above the ground are the two objects when the ball passes the flowepot?**

**2. a = v/t**

all other variations of this constant acceleration formula.

d = v(initial) x t + 1/2 x a x t^2

all other variations of this constant acceleration formula.

d = v(initial) x t + 1/2 x a x t^2

**3. 1. I used the above formula (The one with the 'd') to find the time of flight for the ball.**

2. I used the same formula to find the distance flowerpot fell in 1 second.

3. I subtracted it from the overal distance to find the position of the flowerpot when the ball was dropped.

4. I used that position to find the velocity at that time.

5. Using that 'd' formula again I find the time it takes for the flower pot to fall after the ball is dropped. (Not so sure about this line anymore)

6. No idea how to find the displacement. All I know is that the displacement must equal at that point...

2. I used the same formula to find the distance flowerpot fell in 1 second.

3. I subtracted it from the overal distance to find the position of the flowerpot when the ball was dropped.

4. I used that position to find the velocity at that time.

5. Using that 'd' formula again I find the time it takes for the flower pot to fall after the ball is dropped. (Not so sure about this line anymore)

6. No idea how to find the displacement. All I know is that the displacement must equal at that point...