Two students are on a balcony that is 15 m above the ground. One student throws a ball vertically downward with an initial velocity of 12 m/s. Another student throws a ball upward at the same time with exactly the same speed. Both Balls eventually hit the ground below the balcony.

a.)What order do the balls hit, and how much time elapses between the hits? Ball A hits first and then after 2.4 seconds Ball B hits.
b.)Calculate the final velocity with which each ball hits the ground. A-46.48m/s B-?
c.)What is the distance between the two balls 0.5s after they are thrown? ???

c.) The distance between the two balls 0.5 seconds after they are thrown can be calculated using the formula for vertical motion:

Distance = initial velocity * time + (1/2) * acceleration * time^2

For Ball A, the initial velocity is 12 m/s, the time is 0.5 seconds, and the acceleration due to gravity is -9.8 m/s^2 (taking downward direction as negative). Plugging these values into the formula:

Distance for Ball A = 12 * 0.5 + (1/2) * (-9.8) * (0.5^2)
= 6 + (-0.5) * (-9.8) * 0.25
= 6 + (-0.5) * (-2.45)
= 6 + 1.225
= 7.225 meters

For Ball B, the initial velocity is also 12 m/s (though in the upward direction), the time is 0.5 seconds, and the acceleration due to gravity is -9.8 m/s^2. Plugging these values into the formula:

Distance for Ball B = 12 * 0.5 + (1/2) * (-9.8) * (0.5^2)
= 6 + (-0.5) * (-9.8) * 0.25
= 6 + (-0.5) * (-2.45)
= 6 + 1.225
= 7.225 meters

Therefore, the distance between the two balls 0.5 seconds after they are thrown is 7.225 meters.

To calculate the final velocity for Ball B, we can use the following kinematic equation:

v = u + at

Where:
v = final velocity
u = initial velocity (12 m/s)
a = acceleration (approximately -9.8 m/s^2, since the ball is moving upward against gravity)
t = time (2.4 seconds)

Plugging in the values, we get:

v = 12 + (-9.8)(2.4)
v = 12 - 23.52
v ≈ -11.52 m/s

The negative sign indicates that the ball is moving downward when it hits the ground.

Now, to calculate the distance between the two balls 0.5 seconds after they are thrown, we need to calculate how far each ball has fallen during that time.

For Ball A:
Using the equation:
s = ut + (1/2)at^2

Where:
s = distance (which we need to find)
u = initial velocity (12 m/s)
a = acceleration (approximately 9.8 m/s^2, since the ball is moving downward due to gravity)
t = time (0.5 seconds)

Plugging in the values, we get:

s = (12)(0.5) + (1/2)(9.8)(0.5)^2
s = 6 + (1/2)(9.8)(0.25)
s = 6 + (1/2)(2.45)
s = 6 + 1.225
s ≈ 7.225 m

For Ball B:
Since Ball B is moving upward, we need to calculate the distance it has covered in 0.5 seconds. To do this, we will use the same formula as for Ball A, but with a negative acceleration, since its velocity is decreasing.

s = ut + (1/2)at^2

Where:
s = distance (which we need to find)
u = initial velocity (12 m/s)
a = acceleration (approximately -9.8 m/s^2, since the ball is moving upward against gravity)
t = time (0.5 seconds)

Plugging in the values, we get:

s = (12)(0.5) + (1/2)(-9.8)(0.5)^2
s = 6 + (1/2)(-9.8)(0.25)
s = 6 - (1/2)(2.45)
s = 6 - 1.225
s ≈ 4.775 m

Therefore, the distance between the two balls 0.5 seconds after they are thrown is approximately 7.225 m - 4.775 m = 2.45 m.