A ball thrown horizontally at 22.2 m/s from the roof of a building lands 35.0 m from the base of the building. How high is the building?

I have no idea how to even start this problem. Please help!

Dx = Xo * Tf = 35 m.

22.2 * Tf = 35
Tf = 1.577 s. = Fall time.

h = 0.5g*Tf^2

To solve this problem, you can use the kinematic equations of motion.

In this case, since the ball is thrown horizontally, the initial vertical velocity (Vy) is 0 m/s, while the initial horizontal velocity (Vx) is 22.2 m/s.

The vertical motion of the ball can be modeled using the equation:
h = Viy * t + (1/2) * a * t^2

Since the initial vertical velocity is 0 m/s, this equation simplifies to:
h = (1/2) * a * t^2

The horizontal motion of the ball can be modeled using the equation:
d = Vix * t

Since the ball is thrown horizontally, the initial horizontal velocity remains constant throughout its motion. Thus, the equation simplifies to:
d = Vix * t

Given that the ball lands 35.0 m from the base of the building, we can substitute this value for "d" in the horizontal motion equation:
35.0 m = 22.2 m/s * t

From this equation, we can solve for t:
t = 35.0 m / 22.2 m/s = 1.58 s

Now, we can substitute the value of t into the vertical motion equation to solve for h:
h = (1/2) * a * t^2

The acceleration due to gravity (a) is approximately 9.8 m/s^2:
h = (1/2) * (9.8 m/s^2) * (1.58 s)^2

Calculating h gives:
h = 12.53 m

Therefore, the height of the building is approximately 12.53 meters.

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

First, let's define the variables:
- Initial velocity in the horizontal direction (Vx) = 22.2 m/s
- Distance traveled in the horizontal direction (x) = 35.0 m
- Acceleration in the horizontal direction (ax) = 0 m/s^2
- Time taken to reach the ground (t)

The horizontal motion of the ball is uniform since there is no acceleration in that direction. Therefore, we can use the equation:
x = Vx * t

We need to find the time taken to reach the ground, so let's rearrange the equation:
t = x / Vx

Substituting the given values, we have:
t = 35.0 m / 22.2 m/s ≈ 1.5766 s (rounded to four decimal places)

Now, since the ball was thrown horizontally, the vertical motion is subject to a constant acceleration due to gravity (g = 9.8 m/s^2), and the initial vertical velocity (Vy) is 0 m/s since the ball was thrown horizontally.

We can use the equation of motion for vertical distance to calculate the height of the building:
y = Vy * t + (1/2) * a * t^2

Since Vy = 0 m/s and a = g, the equation simplifies to:
y = (1/2) * g * t^2

Substituting the values, we get:
y = (1/2) * 9.8 m/s^2 * (1.5766 s)^2 ≈ 12.22 m (rounded to two decimal places)

Therefore, the height of the building is approximately 12.22 meters.