https://file.aiquickdraw.com/mathbot/file/1730703676123641p426s.jpg

To solve the problem in the image, we will use the given data and the standard equations of projectile motion. Here's the given information:

- Initial velocity, \( v_0 = 6 \, \text{m/s} \)
- Launch angle, \( \theta_0 = 60^\circ \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)

We need to find:
1. \( R \) (Range)
2. \( t_R \) (Time of flight)
3. \( H \) (Maximum height)
4. \( v_H \) (Horizontal component of velocity at maximum height)

Let's solve each part step-by-step.

### 1. Range (\( R \))
The formula for the range \( R \) of a projectile is:
\[
R = \frac{{v_0^2 \sin(2\theta_0)}}{g}
\]
Substitute the values:
\[
R = \frac{{6^2 \sin(120^\circ)}}{9.81}
\]

Using \( \sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
R = \frac{36 \times \frac{\sqrt{3}}{2}}{9.81}
\]

Calculate \( R \) to find the range.

### 2. Time of flight (\( t_R \))
The time of flight \( t_R \) is given by:
\[
t_R = \frac{2 v_0 \sin(\theta_0)}{g}
\]
Substitute the values:
\[
t_R = \frac{2 \times 6 \times \sin(60^\circ)}{9.81}
\]

Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
t_R = \frac{2 \times 6 \times \frac{\sqrt{3}}{2}}{9.81}
\]

Calculate \( t_R \) to find the time of flight.

### 3. Maximum height (\( H \))
The maximum height \( H \) is given by:
\[
H = \frac{(v_0 \sin(\theta_0))^2}{2g}
\]
Substitute the values:
\[
H = \frac{(6 \times \sin(60^\circ))^2}{2 \times 9.81}
\]

Calculate \( H \) to find the maximum height.

### 4. Horizontal component of velocity at maximum height (\( v_H \))
The horizontal component of velocity \( v_H \) remains constant throughout the motion and is given by:
\[
v_H = v_0 \cos(\theta_0)
\]
Substitute the values:
\[
v_H = 6 \times \cos(60^\circ)
\]

Using \( \cos(60^\circ) = 0.5 \):
\[
v_H = 6 \times 0.5 = 3 \, \text{m/s}
\]

So, \( v_H = 3 \, \text{m/s} \).

Let me know if you would like detailed calculations for each result or if you have any questions.

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### Relative Questions:
1. How is the range of a projectile affected by the launch angle?
2. What would happen to the time of flight if the initial velocity is doubled?
3. How does the angle of projection affect the maximum height achieved?
4. Why does the horizontal velocity component remain constant in projectile motion?
5. How would air resistance change the results for the range and maximum height?

### Tip:
Remember, for projectile motion at an angle, breaking down the initial velocity into horizontal and vertical components makes it easier to solve for various parameters.

A projectile is launched with an initial velocity of 6 m/s at an angle of 60° above the horizontal. Find the range (R), time of flight (t_R), maximum height (H), and horizontal component of velocity at maximum height (v_H) assuming gravitational acceleration is 9.81 m/s².

Learn how to solve a projectile motion problem with an initial velocity of 6 m/s at a 60° angle. Find the range, time of flight, maximum height, and horizontal velocity component. This problem covers key concepts in trigonometry and kinematics, suitable for high school physics students.

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