Two students are on a balcony 22.3 m above the street. One student throws a ball, b1, vertically downward at 19.2 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a) What is the difference in time the balls spend in the air?
s

(b) What is the velocity of each ball as it strikes the ground?
velocity for b1 m/s
velocity for b2 m/s

(c) How far apart are the balls 0.580 s after they are thrown?
m

To solve this problem, we can use the following kinematic equations:

1. Position equation: h = vi*t + (1/2)*a*t^2
2. Velocity equation: v = vi + a*t
3. Final velocity equation: v^2 = vi^2 + 2*a*h

Given:
Initial velocity (vi) for both balls = 19.2 m/s (one upward, one downward)
Height (h) of the balcony = 22.3 m
Time (t) after the balls are thrown = 0.580 s
Acceleration due to gravity (a) = -9.8 m/s^2

Let's break down the problem into parts and solve step-by-step.

(a) Difference in time the balls spend in the air:
Since one ball missed the balcony on the way down, we can find the time it takes for that ball to reach the ground.

For ball 1 (b1) thrown downward:
h1 = vi*t1 + (1/2)*a*t1^2

Using the position equation, we can solve for t1:

22.3 = (19.2)*t1 + (1/2)*(-9.8)*(t1)^2
0 = -4.9*t1^2 + 19.2*t1 - 22.3

Using the quadratic formula, we can find the value of t1:

t1 = (-b ± sqrt(b^2 - 4ac))/(2a)

where a = -4.9, b = 19.2, and c = -22.3

Calculating t1 gives two values: t1 ≈ 0.72 s and t1 ≈ 3.41 s.

Since the second time value is greater than the given time, t1 ≈ 0.72 s will be the correct answer.

Now, we need to find the time for ball 2 (b2) thrown upward to reach the ground.

For ball 2 (b2) thrown upward:
h2 = vi*t2 + (1/2)*a*t2^2

Using the position equation, we can solve for t2:

0 = (19.2)*t2 + (1/2)*(-9.8)*(t2)^2

Applying the quadratic formula, we find only one positive value for t2, which is approximately t2 ≈ 3.91 s.

The difference in time is: t2 - t1 = 3.91 - 0.72 ≈ 3.19 s.

Therefore, the difference in time the balls spend in the air is approximately 3.19 seconds.

(b) Velocity of each ball as it strikes the ground:

For ball 1 (b1) thrown downward:
Using the velocity equation, v = vi + a*t1, we can find the velocity of b1:

v1 = 19.2 + (-9.8)*(0.72)
v1 ≈ 12.96 m/s

For ball 2 (b2) thrown upward:
Using the velocity equation, v = vi + a*t2, we can find the velocity of b2:

v2 = 19.2 + (-9.8)*(3.91)
v2 ≈ -20.52 m/s

The negative sign indicates the ball is moving downward.

Therefore, the velocity of b1 as it strikes the ground is approximately 12.96 m/s, and the velocity of b2 as it strikes the ground is approximately -20.52 m/s.

(c) Distance between the balls after 0.580 s:
To find the distance between the balls after 0.580 s, we need to find the positions of the balls at that time.

For ball 1 (b1) thrown downward:
Using the position equation, h1 = vi*t + (1/2)*a*t^2, we can calculate the position of b1 after 0.580 s:

h1(0.580) = (19.2)*(0.580) + (1/2)*(-9.8)*(0.580)^2
h1(0.580) ≈ 1.459 m

For ball 2 (b2) thrown upward:
Using the position equation, h2 = vi*t + (1/2)*a*t^2, we can calculate the position of b2 after 0.580 s:

h2(0.580) = (19.2)*(0.580) + (1/2)*(-9.8)*(0.580)^2
h2(0.580) ≈ 5.063 m

Therefore, the distance apart between the balls 0.580 s after they are thrown is approximately 5.063 - 1.459 ≈ 3.604 m.

To solve this problem, we need to use the kinematic equations of motion. Let's break down each part of the question.

(a) To find the difference in time the balls spend in the air, we first need to determine the time it takes for each ball to reach the ground.

For ball b1, since it is thrown vertically downward, the acceleration due to gravity will act in the same direction as its initial velocity. The equation we can use to find the time of flight is:

h = (1/2) * g * t^2

where h is the distance traveled, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken. In this case, since the ball is thrown from a height of 22.3 m and we want to find the time it takes to reach the ground, we can rearrange the equation as follows:

22.3 m = (1/2) * (-9.8 m/s^2) * t^2

Solving this equation will give us the time taken for ball b1 to reach the ground.

For ball b2, since it is thrown vertically upward, it will eventually reach its highest point and come back down. The equation for this type of motion is:

v_f = v_i + g * t

where v_f is the final velocity, v_i is the initial velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken. In this case, we know the initial velocity (19.2 m/s) and want to find the time taken to reach the ground. We can rearrange the equation as follows:

0 m/s = 19.2 m/s - 9.8 m/s^2 * t

Solving this equation will give us the time taken for ball b2 to reach the maximum height and subsequently come back down to the ground.

Once we have the time taken for each ball, we can find the difference in time between them.

(b) To find the velocity of each ball as it strikes the ground, we can use the formula:

v_f = v_i + g * t

where v_f is the final velocity, v_i is the initial velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken. For ball b1, the initial velocity is 19.2 m/s downward, and for ball b2, the initial velocity is 19.2 m/s upward. Plugging in the values and solving the equations will give us the velocities at impact.

(c) To find how far apart the balls are 0.580 s after they are thrown, we need to calculate the vertical displacements of each ball. The formula for displacement is:

h = v_i * t + (1/2) * g * t^2

where h is the vertical displacement, v_i is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken. For each ball, we can plug in the values and calculate their displacements. The horizontal distance between the balls will remain constant, so we only need to calculate the vertical distance.

Using these methods, we can solve the problem step by step to find the answers to each part.