Somehow two students managed to open two windows at the library. Student A is 17 m above the ground and Student B is 23.5 m above the ground. At the same time that Student A drops a ball, Student B throws a ball straight down at an initial speed of 7.36 m/s.

At what height above the ground will Student B's ball catch up and pass Student A's ball?
At that moment, what is the difference in speed of the two balls?

Below answer with the height above the ground.

at time t they will be at the same height, when

17-4.9t^2 = 23.5-7.36t-4.9t^2
use that t to get the height

since gravity affects both balls equally, their speed difference is constant.

To find the height at which Student B's ball will catch up and pass Student A's ball, we can start by determining the time it takes for each ball to reach the ground. We can use the kinematic equations of motion to solve for this.

For Student A's ball:
Let's assume the height hA at which Student A's ball catches up with the ground is 0 m. The initial height (hA) is 17 m and the acceleration due to gravity (g) is 9.8 m/s^2.

Using the equation:
hA = h0A + v0At + 1/2gt^2

Substituting the known values:
0 = 17 + 0t + 1/2(9.8)t^2

Simplifying:
4.9t^2 - 17 = 0

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

t = (√(17^2 - 4(4.9)(-17))) / (2(4.9))

t ≈ 2.090 s

Therefore, it takes approximately 2.090 seconds for Student A's ball to reach the ground.

For Student B's ball:
Since Student B throws the ball straight down with an initial speed of 7.36 m/s, and the acceleration due to gravity is still 9.8 m/s^2, we can use the formula for finding the time it takes for an object to reach the ground when thrown vertically upward or downward:

tB = (√(2hB / g))

Substituting the known values:
tB = (√(2(23.5) / 9.8))

tB ≈ 2.181 s

Therefore, it takes approximately 2.181 seconds for Student B's ball to reach the ground.

Since we are looking for the height at which Student B's ball catches up and passes Student A's ball, we need to find the height at that specific time. We can use the equation for the height of an object in freefall:

hB = h0B + v0BtB + 1/2gtB^2

Substituting the known values:
hB = 23.5 + (7.36)(2.181) + 1/2(9.8)(2.181)^2

Simplifying:
hB ≈ 39.366 m

Therefore, at the moment when Student B's ball catches up and passes Student A's ball, it would be at a height of approximately 39.366 meters above the ground.

To find the difference in speed between the two balls at that moment, we can subtract the velocity of Student A's ball from the velocity of Student B's ball.

The velocity of Student A's ball can be calculated using the formula:
vA = v0A + gtA

Substituting the known values:
vA = 0 + (9.8)(2.090)

vA ≈ 20.402 m/s

The velocity of Student B's ball remains constant throughout, since it is thrown straight down. Therefore, the velocity of Student B's ball will be equal to the initial velocity:
vB = 7.36 m/s

The difference in speed between the two balls would be:
vB - vA = 7.36 - 20.402 ≈ -13.042 m/s

Therefore, at that moment, the difference in speed of the two balls would be approximately 13.042 m/s, with Student B's ball having a lower speed.