two students are on a balcony 20.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (Assume the positive direction is upward.)

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

(b) What is the velocity of each ball as it strikes the ground?
ball thrown downward
m/s
ball thrown upward
m/s

(c) How far apart are the balls 0.800 s after they are thrown?

t=.378s

Vf=18.4m/s

To solve this problem, we'll use the equations of motion to find the necessary information.

(a) Let's start by finding the time it takes for the first ball to reach the ground. We can use the equation:

h = (1/2)gt^2

Where h is the height, g is the acceleration due to gravity, and t is the time.

For the first ball, h = 20.6 m and g = 9.8 m/s^2. Plugging in these values, we can solve for the time:

20.6 = (1/2)(9.8)t^2
41.2 = 9.8t^2
t^2 = 41.2/9.8
t^2 = 4.2
t = sqrt(4.2)
t ≈ 2.05 s

Now let's find the time it takes for the second ball to reach the ground. Since the second ball just misses the balcony on the way down, it will have the same time of flight as the first ball. Therefore, the time for the second ball is also approximately 2.05 s.

The difference in the time the balls spend in the air is the time taken by the first ball minus the time taken by the second ball:

Time difference = 2.05 s - 2.05 s = 0 s

Therefore, the difference in the time the balls spend in the air is 0 seconds.

(b) Since the first ball is thrown downward, its velocity when it strikes the ground will be positive. We can use the equation:

v = u + gt

Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

For the first ball, u = 14.7 m/s (since it's thrown downward) and g = 9.8 m/s^2. Plugging in these values and using the time of flight we found earlier (2.05 s):

v = 14.7 + (9.8)(2.05)
v ≈ 14.7 + 20
v ≈ 34.7 m/s

Therefore, the velocity of the first ball as it strikes the ground is approximately 34.7 m/s.

For the second ball, since it's thrown upward, its velocity will be negative when it strikes the ground. We can use the same equation, but with u = -14.7 m/s:

v = -14.7 + (9.8)(2.05)
v ≈ -14.7 + 20
v ≈ 5.3 m/s

Therefore, the velocity of the second ball as it strikes the ground is approximately 5.3 m/s.

(c) To find how far apart the balls are 0.800 s after they are thrown, we need to calculate the distance traveled by each ball in that time.

For the first ball, we can use the equation:

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

Where s is the distance, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

For the first ball, u = 14.7 m/s and t = 0.800 s:

s = (14.7)(0.800) + (1/2)(9.8)(0.800)^2
s ≈ 11.76 + 3.92
s ≈ 15.68 m

Therefore, the first ball travels approximately 15.68 m in 0.800 s.

For the second ball, we can use the same equation, but with u = -14.7 m/s:

s = (-14.7)(0.800) + (1/2)(9.8)(0.800)^2
s ≈ -11.76 + 3.92
s ≈ -7.84 m

Therefore, the second ball travels approximately -7.84 m in 0.800 s.

The distance between the balls 0.800 s after they are thrown is the absolute difference between their distances:

Distance = |15.68 - (-7.84)|
Distance = |15.68 + 7.84|
Distance = 23.52 m

Therefore, the balls are approximately 23.52 m apart 0.800 s after they are thrown.