2 tall buildings are 200m apart. with what speed a ball be thrown horizontally from a window of one building 2km above the ground so that it will enter a window 40m from the ground in the other?

2000 meters - 40 meters = 1960 meter fall

vertical fall problem first
1960 = 4.9 t^2
t = 20 second fall

so it has to go 200 meters horizontal in 20 seconds
u = 200/20 = 10 m/s

To find the speed at which the ball needs to be thrown horizontally, we can use the equations of motion.

Let's break down the information given in the problem:
1. The two tall buildings are 200m apart horizontally.
2. The ball is thrown from a window of one building, which is 2km (2000m) above the ground.
3. The ball needs to enter a window on the other building, which is 40m above the ground.

First, let's calculate the time it takes for the ball to reach the other building (the horizontal distance remains constant). We can use the equation:

d = v * t

where d is the distance, v is the velocity, and t is the time.

Since the ball is thrown horizontally, its initial vertical velocity is zero. As a result, the only force acting on the ball is gravity, causing it to fall vertically. The time taken for a body to fall freely from a given height can be calculated using the equation:

h = (1/2) * g * t^2

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Using this equation, we can calculate the time it takes for the ball to fall from a height of 2km (2000m):

2000m = (1/2) * 9.8m/s^2 * t^2

Simplifying the equation:

t^2 = (2 * 2000m) / 9.8m/s^2
t^2 = 4000m / 9.8m/s^2
t^2 ≈ 408.16s^2
t ≈ √408.16s^2
t ≈ 20.2s (approx.)

Now that we have the time taken for the vertical motion, we can use this value to calculate the horizontal velocity.

Using the equation:

v = d / t

where v is the velocity, d is the distance, and t is the time.

The distance between the two buildings is given as 200m, and we've calculated the time t as 20.2s. So, we can calculate the horizontal velocity:

v = 200m / 20.2s
v ≈ 9.90 m/s

Therefore, the ball needs to be thrown horizontally from the window with a speed of approximately 9.90 m/s to enter the other window 40m above the ground.