The Falling Paint Can: A painter is standing on scaffolding that is raised at a constant speed vi. As he travels upward he accidentally nudges a paint can off the scaffolding and it falls 15m to the ground. You are watching and measure with your stopwatch that it takes 3.25s for the can to reach the ground.

a.What is the speed of the can just before it hits the ground?
b.Another painter is standing on the ledge, which his hands 4m above the can when it falls off. He has lightning fast reflexes and if the paint can passes in front of him, he can catch it. Does he get the chance?

a. In order to find the speed of the paint can just before it hits the ground, we can use the equation:

v_f = v_i + a*t

where v_f is the final velocity, v_i is the initial velocity (0 m/s in this case, since the can is moving upward with the same speed as the scaffolding), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken for the fall (3.25 seconds).

v_f = 0 + (-9.8) * 3.25
v_f = -31.85 m/s

The speed of the can just before it hits the ground is 31.85 m/s downward.

b. To find out if the other painter has a chance to catch the can, we first need to find the distance the can falls during the time he has to react. We will assume the painter has an instant reaction time.

We can use the equation:

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

where h is the distance fallen, v_i is the initial velocity, a is the acceleration, and t is the time taken.

h = 0 * 3.25 + 0.5 * (-9.8) * 3.25^2
h = -51.53 m

Since the distance the paint can falls (-51.53 m) is greater than the height of the other painter's hands (4 m above the can when it falls off, so 15 + 4 = 19 m), it will pass in front of him, giving him a chance to catch it.

a. The speed of the can just before it hits the ground can be found using the equation v = d/t, where v is the speed, d is the distance, and t is the time. In this case, the distance is 15m and the time is 3.25s. Plugging in the values, we get v = 15m/3.25s ≈ 4.615m/s. So, the speed of the can just before it hits the ground is approximately 4.615m/s.

b. The second painter, with lightning-fast reflexes, has his hands 4m above the can. Since the can falls from a height of 15m, and the second painter is only 4m above it, he won't get the chance to catch it. The can will fall past him before he can react. Looks like this time his superpowers won't be of much use!

a. To find the speed of the can just before it hits the ground, we can use the formula for the final velocity of an object in free fall:

v = vi + gt

Where:
v = final velocity of the can
vi = initial velocity of the can (0 m/s since it was released from rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time taken to reach the ground (3.25 s)

Substituting the given values into the formula:

v = 0 + (-9.8 m/s^2)(3.25 s)
v = -31.85 m/s

The negative sign indicates that the velocity is in the downward direction. So, the speed of the can just before it hits the ground is approximately 31.85 m/s.

b. The painter on the ledge will have a chance to catch the can if he can reach it before it hits the ground. To determine this, we can calculate the time it takes for the can to fall 4m (height of the painter's hands above the can).

Using the equation for displacement in free fall:

y = vi*t + (1/2)gt^2

Where:
y = displacement (4m)
vi = initial velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken to fall 4m (unknown)

Substituting the given values into the equation:

4 = 0*t + (1/2)(-9.8 m/s^2)t^2
2(t^2) = 4
t^2 = 2
t = √2 (√ denotes square root)

t ≈ 1.41 s

Since the time taken for the can to reach the ground is 3.25s, and it takes 1.41s to fall 4m, the painter on the ledge will have enough time to catch the can before it hits the ground.

To find the answers to both questions, we'll need to use kinematic equations and the laws of motion. Let's break down the problem step by step.

a. To find the speed of the can just before it hits the ground, we can start by using the equation for the distance covered by an object in free fall:

d = vi * t + (1/2) * g * t^2

In this equation, d represents the distance (15m), vi represents the initial velocity (unknown), t represents the time (3.25s), and g represents the acceleration due to gravity (approximately 9.8m/s^2).

We need to solve for vi. Rearranging the equation:

15m = vi * 3.25s + (1/2) * 9.8m/s^2 * (3.25s)^2

Now, we can solve for vi:

vi * 3.25s = 15m - (1/2) * 9.8m/s^2 * (3.25s)^2

vi * 3.25s = 15m - 51.609375m

vi * 3.25s = -36.609375m

vi = -36.609375m / 3.25s

vi ≈ -11.253m/s (rounded to three decimal places)

The speed of the can just before it hits the ground is approximately 11.253 m/s (rounded to three decimal places). Note that the negative sign indicates the direction of motion (downward).

b. To determine whether the second painter can catch the can, we need to calculate the time it takes for the can to fall 4m from the ledge to the ground.

Using the same equation as before, d = vi * t + (1/2) * g * t^2, we substitute d = 4m, vi = unknown, g = 9.8m/s^2, and solve for t.

4m = vi * t + (1/2) * 9.8m/s^2 * t^2

Since we know the time it takes for the can to reach the ground is 3.25s, we need to determine if this time is greater or lesser than the time it takes to fall 4m.

Now, we solve the equation:

4m = vi * 3.25s + (1/2) * 9.8m/s^2 * (3.25s)^2

4m = 3.25vi + 52.78125

3.25vi = 4m - 52.78125

3.25vi = -48.78125

vi = -48.78125 / 3.25

vi ≈ -15.019m/s (rounded to three decimal places)

The speed of the can just before reaching a height of 4m (the ledge) is approximately 15.019 m/s (rounded to three decimal places). Again, the negative sign indicates the downward motion.

Since the initial velocity of the can at the ledge is greater than its speed just before hitting the ground, the second painter does not get the chance to catch the can.