A football is kicked at ground level with a speed of 20.0 m/s at an angle of 37.0 degrees to the horizontal. How much later does it hit the ground? How far down from the field does the ball go?

To solve this problem, we can treat it as a projectile motion question where a football is kicked at an initial velocity \( v_0 = 20.0 \, \text{m/s} \) and at an angle \( \theta = 37.0^\circ \) to the horizontal. We are tasked with finding two things:

1. The time the ball takes to hit the ground (i.e., total flight time).
2. The horizontal distance the ball travels (i.e., the range).

### Step 1: Break the Initial Velocity into Components
We need to resolve the initial velocity into its horizontal and vertical components using trigonometry:

\[
v_{0x} = v_0 \cos(\theta)
\]
\[
v_{0y} = v_0 \sin(\theta)
\]

Where:
- \( v_0 = 20.0 \, \text{m/s} \)
- \( \theta = 37.0^\circ \)

Calculating the components:
\[
v_{0x} = 20.0 \, \text{m/s} \times \cos(37^\circ) = 20.0 \times 0.7986 = 15.97 \, \text{m/s}
\]
\[
v_{0y} = 20.0 \, \text{m/s} \times \sin(37^\circ) = 20.0 \times 0.6018 = 12.04 \, \text{m/s}
\]

### Step 2: Time of Flight
The total time of flight can be determined by analyzing the vertical motion. The ball will rise to a maximum height and then fall back to the ground. The equation for vertical displacement is:

\[
y = v_{0y} t - \frac{1}{2} g t^2
\]

Since the ball starts and ends at ground level, \( y = 0 \). Thus, the equation becomes:

\[
0 = v_{0y} t - \frac{1}{2} g t^2
\]

Factoring out \( t \) gives:

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

This equation has two solutions:
1. \( t = 0 \) (the time when the ball is kicked).
2. \( t = \frac{2v_{0y}}{g} \) (the total time of flight).

Substituting the values for \( v_{0y} \) and \( g = 9.81 \, \text{m/s}^2 \):

\[
t = \frac{2 \times 12.04 \, \text{m/s}}{9.81 \, \text{m/s}^2} = \frac{24.08}{9.81} = 2.45 \, \text{seconds}
\]

### Step 3: Horizontal Range
The horizontal range (distance) can be found using the horizontal velocity and the total time of flight:

\[
\text{Range} = v_{0x} \times t
\]

Substituting the values:

\[
\text{Range} = 15.97 \, \text{m/s} \times 2.45 \, \text{seconds} = 39.15 \, \text{m}
\]

### Final Answers:
1. **Time of flight**: 2.45 seconds
2. **Range (horizontal distance)**: 39.15 meters

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the maximum height the ball reaches?
2. How would the time of flight change if the ball were kicked at a different angle?
3. What happens to the horizontal range if the initial velocity is increased to 25 m/s?
4. Can air resistance affect the time of flight and range in a real-world scenario?
5. How would the problem change if the ball were kicked from an elevated platform?

**Tip**: In projectile motion, the horizontal velocity remains constant if air resistance is neglected, while the vertical velocity changes due to gravity.

Calculate the time a football stays in the air and how far it travels when kicked at 20.0 m/s at a 37-degree angle. Learn about projectile motion concepts, kinematic equations, and the role of trigonometry. Suitable for Grades 10-12 physics students.

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