A man holds a flower pot out a window 12m from the ground. He tosses the pot upward with a speed of 5m/s. How long does it take the pot to reach the ground? How fast is it moving just before it hits? Please show steps.

To find the time it takes for the pot to reach the ground, we can use the equation of motion:

h = ut + (1/2)gt^2

where:
h = height (12m)
u = initial velocity (5m/s)
g = acceleration due to gravity (-9.8m/s^2)
t = time

Since the pot is tossed upward, its initial velocity (u) is positive, but the acceleration due to gravity (g) is negative because it acts in the opposite direction. The final height (h) is zero since it reaches the ground.

Plugging in the values, we have:

0 = 5t + (1/2)(-9.8)t^2

Simplifying the equation:

0 = 5t - 4.9t^2

Rearranging the equation:

4.9t^2 - 5t = 0

Factoring out t:

t(4.9t - 5) = 0

This equation can be satisfied if either t = 0 (which we discard since it represents the initial time) or (4.9t - 5) = 0.

Solving (4.9t - 5) = 0:

4.9t = 5

t = 5 / 4.9

t ≈ 1.02 seconds

Therefore, it takes approximately 1.02 seconds for the pot to reach the ground.

To find the speed just before it hits the ground, we need to find the velocity at the time t = 1.02 seconds. We can use the equation:

v = u + gt

where:
v = final velocity
u = initial velocity (5m/s)
g = acceleration due to gravity (-9.8m/s^2)
t = time (1.02 seconds)

Plugging in the values:

v = 5 + (-9.8)(1.02)

v ≈ -4.996

The negative sign indicates that the pot is moving downward just before it hits. Therefore, the speed of the pot just before it hits the ground is approximately 4.996 m/s, downward (or equivalently, upward).