Two students are on a balcony 24.9 m above the street. One student throws a ball, b1, vertically downward at 19.6 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?


(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 0.440 s after they are thrown?

vertical equation.

hf=ho+Vi*t-1/2 g t^2
-24.9 = Vi*t-1/2 g t^2 use the quadratic eqution.
solve for t, given Vi is +, then -
then subtract the difference.

b. Vf=Vi+gt

To solve this problem, we will use the kinematic equations of motion.

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

For ball b1, which is thrown downward:
We can use the equation:

h = (1/2)gt^2

where h is the initial height (24.9 m) and g is the acceleration due to gravity (-9.8 m/s^2). We solve for t:

24.9 = (1/2)(-9.8)t^2
t^2 = (24.9 * 2) / (-9.8)
t^2 = 5.08
t = √5.08
t ≈ 2.25 s

For ball b2, which is thrown upward:
The initial velocity of b2 is 19.6 m/s, and it will take the same amount of time as b1 to reach the highest point of its trajectory. At this point, its velocity will be 0 m/s, and it will start falling. The total time of flight is twice the time it takes to reach the highest point.

So, the total time of flight for ball b2 is 2 * 2.25 s = 4.5 s.

The difference in time the balls spend in the air is 4.5 s - 2.25 s = 2.25 s.

(b) To find the velocity of each ball as it strikes the ground, we need to use the equation:

v = u + gt

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

For ball b1:
The initial velocity is 19.6 m/s (downward), and the time is 2.25 s.
v = 19.6 - 9.8 * 2.25
v ≈ -18.1 m/s (negative sign indicates downward direction)

For ball b2:
The initial velocity is 19.6 m/s (upward), and the time is 4.5 s.
v = 19.6 - 9.8 * 4.5
v ≈ -31.2 m/s (negative sign indicates downward direction)

(c) To find how far apart the balls are 0.440 s after they are thrown, we need to calculate the distance traveled by each ball during this time.

For ball b1:
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 (-9.8 m/s^2), and t is the time.

s1 = 19.6 * 0.44 + (1/2) * (-9.8) * (0.44)^2
s1 ≈ 8.63 m

For ball b2:
The initial velocity is 19.6 m/s, but it's moving in the opposite direction. So, we take the negative of the initial velocity:
u2 = -19.6 m/s

s2 = u2 * 0.44 + (1/2) * (-9.8) * (0.44)^2
s2 ≈ -2.13 m

The distance between the balls is the sum of the distance traveled by each ball:
s_total = s1 + s2
s_total ≈ 8.63 m + (-2.13 m)
s_total ≈ 6.5 m

Therefore, the balls are approximately 6.5 meters apart 0.440 seconds after they are thrown.