10. A flowerpot is dropped from the balcony of an apartment,

Question

28.5 m above the ground. At a time of 1.00 s after the pot is

dropped, a ball is thrown vertically downward from the bal-
cony 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 flowerpot?

Answer 1

To find out if the ball passes the flowerpot before striking the ground, we first need to calculate the time it takes for the ball to reach the ground. We can use the following kinematic equation for the ball's vertical motion:

d = v_i * t + 0.5 * a * t^2

Where:
d = distance traveled
v_i = initial velocity
t = time elapsed
a = acceleration (in this case, due to gravity, approximated as -9.8 m/s^2)

Solving for t:

0 = 12.0 * t - 0.5 * 9.8 * t^2

Rearranging the equation:

4.9t^2 - 12t = 0

Factoring out t:

t(4.9t - 12) = 0

So, either t = 0 (initial time) or 4.9t - 12 = 0

Let's solve for t using the second equation:

4.9t - 12 = 0
4.9t = 12
t = 12/4.9
t ≈ 2.45 seconds

Now, let's calculate how far above the ground the ball and the flowerpot are at t = 1.00 second.

For the ball:
d_ball = v_i * t + 0.5 * a * t^2
d_ball = 12.0 * 1.00 + 0.5 * 9.8 * 1.00^2
d_ball = 12.0 + 0.5 * 9.8 * 1.00
d_ball = 12.0 + 4.9
d_ball ≈ 16.9 meters

For the flowerpot:
d_pot = v_i * t + 0.5 * a * t^2
d_pot = 0 * 1.00 + 0.5 * 9.8 * 1.00^2
d_pot = 0 + 0.5 * 9.8 * 1.00
d_pot = 0 + 4.9
d_pot = 4.9 meters

Therefore, when t = 1.00 second, the ball is about 16.9 meters above the ground, and the flowerpot is about 4.9 meters above the ground. So, the ball passes the flowerpot before striking the ground.