a long Jumper leaves the ground with an Initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the peak height of the long-Jumper, and the horizontal distance

To solve this problem, we can break the initial velocity into its horizontal and vertical components, then use kinematic equations to find the time of flight, peak height, and horizontal distance.

### Given:
- Initial velocity, \( v_0 = 12 \, \text{m/s} \)
- Angle of projection, \( \theta = 28^\circ \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

### 1. **Time of Flight (T):**

The time of flight is the total time the projectile (in this case, the long jumper) is in the air.

The vertical component of the initial velocity is given by:
\[
v_{0y} = v_0 \sin \theta
\]
The time to reach the peak is when the vertical velocity becomes zero. This can be found using the equation:
\[
v_{y} = v_{0y} - g t
\]
At the peak, \( v_y = 0 \), so:
\[
0 = v_0 \sin \theta - g t_{\text{up}}
\]
\[
t_{\text{up}} = \frac{v_0 \sin \theta}{g}
\]
Since the time to ascend and descend are equal, the total time of flight is:
\[
T = 2 \times t_{\text{up}} = \frac{2 v_0 \sin \theta}{g}
\]

### 2. **Peak Height (H):**

The peak height is the maximum vertical distance the jumper reaches above the ground.

Using the vertical component of the velocity:
\[
H = v_{0y} t_{\text{up}} - \frac{1}{2} g t_{\text{up}}^2
\]
Substituting \( t_{\text{up}} = \frac{v_0 \sin \theta}{g} \) into the height equation:
\[
H = \frac{(v_0 \sin \theta)^2}{2g}
\]

### 3. **Horizontal Distance (Range, R):**

The horizontal distance (range) is the distance traveled by the jumper in the horizontal direction during the time of flight.

The horizontal component of the initial velocity is:
\[
v_{0x} = v_0 \cos \theta
\]
The range is then given by:
\[
R = v_{0x} \times T = v_0 \cos \theta \times \frac{2 v_0 \sin \theta}{g}
\]
This can be simplified to:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]

Now, let's calculate each value using the given data.

### 1. **Time of Flight (T):**
\[
T = \frac{2 \times 12 \times \sin(28^\circ)}{9.81} \approx 1.15 \, \text{seconds}
\]

### 2. **Peak Height (H):**
\[
H = \frac{(12 \times \sin(28^\circ))^2}{2 \times 9.81} \approx 1.61 \, \text{meters}
\]

### 3. **Horizontal Distance (R):**
\[
R = \frac{12^2 \times \sin(2 \times 28^\circ)}{9.81} \approx 12.23 \, \text{meters}
\]

### Summary:
- **Time of Flight (T):** Approximately \( 1.15 \, \text{seconds} \)
- **Peak Height (H):** Approximately \( 1.61 \, \text{meters} \)
- **Horizontal Distance (R):** Approximately \( 12.23 \, \text{meters} \)

Would you like further details on any of these calculations or have any questions?

### Related Questions:
1. How would the results change if the initial velocity were 15 m/s?
2. What would be the time of flight if the angle of projection was increased to 45 degrees?
3. How does air resistance affect the time of flight and horizontal distance?
4. What would be the maximum height if the jumper's angle was increased to 35 degrees?
5. How would the results differ if the acceleration due to gravity was 1.6 m/s², like on the moon?

### Tip:
Remember, the range of a projectile is maximum when the angle of projection is 45 degrees, given the same initial velocity.

Solve a projectile motion problem for a long jumper starting with an initial velocity of 12 m/s at 28 degrees above the horizontal. Calculate the time of flight, peak height, and horizontal distance using kinematic equations and trigonometry. Time of flight is approximately 1.15 seconds, peak height is about 1.61 meters, and horizontal distance is around 12.23 meters.

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