A stone is thrown vertically upward from outside a window, with an initial velocity of 24.0 m/s. It moves up, and then down to the ground 420 m below the point where it started. Let `up' be positive.

(a) What is the maximum height above the window to which the stone rises?
________m.

(b) How long is the rock in the air before it reaches the ground?
________ s.

(c) What is its velocity just before it hits the ground?
_______ m/s.

To solve this problem, we can use the equations of motion for vertical motion.

(a) To find the maximum height above the window, we can use the following equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

In this case, the stone is thrown vertically upward, so the initial velocity (u) is +24.0 m/s and the final velocity (v) is 0 m/s at the highest point where the stone changes direction. The acceleration (a) is due to gravity and is equal to -9.8 m/s^2 since it acts downward. The displacement (s) is the maximum height above the window, which is what we want to find.

Plugging these values into the equation, we have:

0^2 = (24.0)^2 + 2*(-9.8)*s

Simplifying the equation, we get:

0 = 576 - 19.6s

19.6s = 576

s = 576 / 19.6

s ≈ 29.4 m

Therefore, the maximum height above the window to which the stone rises is approximately 29.4 meters.

(b) To find the time it takes for the rock to reach the ground, we can use the following equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Again, since the stone is thrown vertically upward, the initial velocity (u) is +24.0 m/s and the final velocity (v) is -24.0 m/s when it reaches the ground. The acceleration (a) is -9.8 m/s^2. We want to find the time (t) it takes for the stone to reach the ground.

Plugging these values into the equation, we have:

-24.0 = 24.0 + (-9.8)t

-48.0 = -9.8t

t = -48.0 / -9.8

t ≈ 4.9 s

Therefore, the stone is in the air for approximately 4.9 seconds before it reaches the ground.

(c) To find the velocity just before the stone hits the ground, we can use the following equation:

v = u + at

Again, since the stone is thrown vertically upward, the initial velocity (u) is +24.0 m/s and the acceleration (a) is -9.8 m/s^2. We want to find the final velocity (v) just before the stone hits the ground.

Plugging these values into the equation, we have:

v = 24.0 + (-9.8)(4.9)

v ≈ -24.0 + (-48.02)

v ≈ -72.0

Therefore, the velocity just before the stone hits the ground is approximately -72.0 m/s. The negative sign indicates that the velocity is directed downward.