A large rock is dropped from a plane flying horizontally at an altitude of 4.00 x 10 ^3 ( to the power of 3) and with a speed of 900km h.

a) calculate the distance that the rock will move horizontally before hitting the ground

b) find the velocity of the rock just before it hits the ground

900km/hr = 250m/s

How long does it take to fall 4000m?
4.9t^2 = 4000
t = 28.57s

How far does the rock travel?
250m/s * 28.57s = 7142.8m = 7.1km

What's its speed?
horizontal: 250m/s
vertical: 9.8*28.57=280m/s
total speed: √(250^2 + 280^2) = 375.4m/s

To calculate the distance that the rock will move horizontally before hitting the ground, we need to determine the time it takes for the rock to reach the ground.

a) Let's start by converting the speed of the plane from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. Therefore, the speed of the plane is 900 km/h x 0.2778 m/s = 250 m/s.

Next, we can calculate the time it takes for the rock to fall to the ground. It is important to note that the vertical motion of the rock (falling vertically due to gravity) is independent of the horizontal motion of the plane.

We can use the equation: distance = initial velocity x time + 0.5 x acceleration x time^2

Since the rock is only falling vertically, the initial velocity (u) is equal to 0, as it was initially at rest.

Therefore, the equation simplifies to: distance = 0.5 x acceleration x time^2

The acceleration due to gravity is approximately 9.8 m/s^2.

Now, we need to calculate the time it takes for the rock to fall. We can use the equation: distance = 0.5 x acceleration x time^2 and substitute the values we know:

4.00 x 10^3 m = 0.5 x 9.8 m/s^2 x time^2

Rearranging the equation, we get:

time^2 = (2 x distance) / acceleration

time^2 = (2 x 4.00 x 10^3 m) / 9.8 m/s^2

time^2 = 816.33 seconds^2

time = √816.33 seconds

time ≈ 28.57 seconds

Now that we know the time it takes for the rock to fall, we can calculate the horizontal distance it will move by multiplying the horizontal speed of the plane (250 m/s) by the time it takes for the rock to fall (28.57 seconds).

Distance = speed x time

Distance = 250 m/s x 28.57 s

Distance ≈ 7142.86 meters

Therefore, the rock will move horizontally approximately 7142.86 meters before hitting the ground.

b) To find the velocity of the rock just before it hits the ground, we can use the formula: velocity = initial velocity + acceleration x time.

Since the rock was initially at rest, the initial velocity (u) is 0. And we know the acceleration due to gravity is approximately 9.8 m/s^2.

Using the time we calculated earlier (28.57 seconds), we can substitute the values into the equation:

velocity = 0 + (9.8 m/s^2) x 28.57 s

velocity = 280 m/s

Therefore, the velocity of the rock just before it hits the ground is 280 m/s.