A stone from a window with an initial horizontal velocity of 10m/s. If the window is 20m high, and the ground is level, how many seconds will the stone reach the ground? How far on the ground will it reach?

a. h = 0.5g*t^2.

h = 20 m., g = 9.8 m/s^2.
t = ?.

b. d = 10m/s * t.
t = Value calculated in part "a".

To find the time it takes for the stone to reach the ground and the distance it will travel on the ground, we need to use the laws of motion and kinematics.

1. Find the time it takes for the stone to reach the ground:
The vertical motion of the stone is influenced by gravity. We can use the equation of motion: h = ut + (1/2)gt^2, where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

Given:
- Initial vertical velocity (u) = 0 (since the stone is not thrown upwards)
- Height (h) = 20m
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)

Using the equation, we can rearrange it to find the time:
h = (1/2)gt^2
20 = (1/2)(9.8)t^2
40 = 9.8t^2
t^2 = 40/9.8
t^2 ≈ 4.08
t ≈ √4.08
t ≈ 2.02 seconds

Therefore, it will take approximately 2.02 seconds for the stone to reach the ground.

2. Find the distance the stone will travel on the ground:
The horizontal motion of the stone is not affected by gravity, so the initial horizontal velocity remains constant.

Given:
- Initial horizontal velocity (u) = 10m/s
- Time (t) = 2.02 seconds (from the previous calculation)

To find the distance (d) traveled horizontally, we use the formula: d = ut
d = 10 * 2.02
d ≈ 20.2 meters

Therefore, the stone will travel approximately 20.2 meters on the ground.