1. A projectile is fired horizontally at an angle of 45° with the speed of 0.075km/s. Find the range and height. The final answer should be in meters.

Given:

Solution:

To solve the problem, we need to break down the motion of the projectile into its horizontal and vertical components. Since the projectile is fired at an angle, we can use trigonometric functions to find these components.

### Given:
- Angle of projection (\(\theta\)): \(45^\circ\)
- Initial speed (\(v_0\)): \(0.075 \, \text{km/s}\) or \(75 \, \text{m/s}\)
- Acceleration due to gravity (\(g\)): \(9.8 \, \text{m/s}^2\)

### Solution:

#### 1. Horizontal and Vertical Components of Initial Velocity:
Since the angle of projection is \(45^\circ\), the horizontal and vertical components of the velocity are equal.

\[
v_{0x} = v_0 \cdot \cos(45^\circ)
\]
\[
v_{0y} = v_0 \cdot \sin(45^\circ)
\]

Using \(\cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}}\):

\[
v_{0x} = v_{0y} = 75 \times \frac{1}{\sqrt{2}} = 75 \times \frac{\sqrt{2}}{2} \approx 53.03 \, \text{m/s}
\]

#### 2. Time of Flight (\(T\)):
The time of flight can be found using the vertical motion equation. At the peak, the vertical velocity becomes zero, and the time to reach the peak is:

\[
t_{\text{peak}} = \frac{v_{0y}}{g} = \frac{53.03}{9.8} \approx 5.41 \, \text{seconds}
\]

Since the time to go up is the same as the time to come down, the total time of flight is:

\[
T = 2 \times t_{\text{peak}} = 2 \times 5.41 \approx 10.82 \, \text{seconds}
\]

#### 3. Range (\(R\)):
The horizontal range can be found by multiplying the horizontal component of the velocity by the time of flight:

\[
R = v_{0x} \times T = 53.03 \times 10.82 \approx 574 \, \text{meters}
\]

#### 4. Maximum Height (\(H\)):
The maximum height can be found using the following kinematic equation:

\[
H = \frac{v_{0y}^2}{2g} = \frac{(53.03)^2}{2 \times 9.8} \approx \frac{2818.16}{19.6} \approx 144 \, \text{meters}
\]

### Final Answer:
- **Range (R):** \(574 \, \text{meters}\)
- **Maximum Height (H):** \(144 \, \text{meters}\)

Would you like further clarification or details on any step?

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Here are five additional questions you might explore:

1. How would the range and height change if the angle were increased to \(60^\circ\)?
2. What would be the effect on the range if the initial speed were doubled?
3. How would air resistance affect the results, and how could it be accounted for?
4. What is the significance of the angle of \(45^\circ\) in projectile motion?
5. How would the maximum height change if the acceleration due to gravity were different (e.g., on the Moon)?

**Tip:** When working with projectile motion problems, always resolve the initial velocity into its horizontal and vertical components first, as this simplifies the calculations.

Learn how to solve a projectile motion problem where a projectile is fired horizontally at a 45° angle with a speed of 0.075 km/s. Find the range and maximum height in meters using trigonometric and kinematic equations.

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