Course Contents » Set 02 (09/18 Wed 10 PM) » Soccer ball on the roof
A soccer ball is kicked from the top of one building with a height of H1 = 31.5 m to another building with a height of H2 = 13.4 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)

The ball is kicked with a speed of v0 = 15.1 m/s at angle of θ = 72.0° with respect to the horizontal. How much time will the ball spend in the air before it hits the roof of the other building? (The soccer ball is kicked without a spin. Neglect air resistance.)


We can solve this problem using the equations of projectile motion, taking into account the vertical and horizontal components of the motion separately. Here's the step-by-step approach:

### Step 1: Break the Initial Velocity into Components
The ball is kicked with an initial speed \( v_0 = 15.1 \, \text{m/s} \) at an angle \( \theta = 72.0^\circ \) above the horizontal. We can find the horizontal and vertical components of the initial velocity:

- Horizontal component: 
\[
v_{0x} = v_0 \cdot \cos(\theta)
\]
- Vertical component:
\[
v_{0y} = v_0 \cdot \sin(\theta)
\]

### Step 2: Use the Kinematic Equation for Vertical Motion
The motion in the vertical direction is governed by the following kinematic equation:

\[
y(t) = y_0 + v_{0y}t - \frac{1}{2} g t^2
\]
Where:
- \( y(t) \) is the final vertical position (the height of the second building, \( H_2 = 13.4 \, \text{m} \)),
- \( y_0 \) is the initial vertical position (the height of the first building, \( H_1 = 31.5 \, \text{m} \)),
- \( v_{0y} \) is the initial vertical velocity component,
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t \) is the time in the air.

Plugging in the known values into this equation, we get:

\[
13.4 = 31.5 + (15.1 \cdot \sin(72.0^\circ))t - \frac{1}{2} (9.81) t^2
\]

Now we solve for \( t \).

### Step 3: Solve the Quadratic Equation
Rearranging the equation into standard quadratic form:

\[
- \frac{1}{2} g t^2 + v_{0y} t + (H_1 - H_2) = 0
\]

Substitute the known values:

\[
-4.905 t^2 + (15.1 \cdot \sin(72.0^\circ))t + (31.5 - 13.4) = 0
\]

This becomes:

\[
-4.905 t^2 + (15.1 \cdot 0.951)t + 18.1 = 0
\]

Now let's solve this quadratic equation for \( t \). I'll calculate the value of \( t \) next.The two possible solutions for the time \( t \) are:

- \( t_1 = -0.95 \, \text{seconds} \) (which is not physically meaningful since time cannot be negative),
- \( t_2 = 3.88 \, \text{seconds} \).

Thus, the ball will spend approximately **3.88 seconds** in the air before hitting the roof of the second building.

Do you want more details or have any questions?

Here are 5 related questions to expand on this concept:
1. What would be the horizontal distance the ball travels before hitting the second building?
2. How would the time of flight change if the initial velocity were increased?
3. What would be the maximum height the ball reaches during its trajectory?
4. How would the trajectory change if there were air resistance?
5. What angle of launch would maximize the horizontal distance?

**Tip:** To find how far an object travels horizontally, use the horizontal component of the velocity and multiply it by the time in the air.

A soccer ball is kicked from the top of one building with a height of H1 = 31.5 m to another building with a height of H2 = 13.4 m. The ball is kicked with a speed of v0 = 15.1 m/s at an angle of θ = 72.0° with respect to the horizontal. How much time will the ball spend in the air before it hits the roof of the other building? (The soccer ball is kicked without a spin. Neglect air resistance.)

Find out how to calculate the time of flight for a soccer ball kicked from a building 31.5 m high to another building 13.4 m high. Using projectile motion and quadratic equations, this problem is solved step-by-step, exploring velocity components, kinematic equations, and solving for time. Suitable for high school students (Grades 10-12).

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