Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

A gun is fired with a velocity of [tex]$250 \, \text{m/s}$[/tex] and attains a maximum height which is one-third of the range. Calculate the following:

i. The angle at which the bullet is fired.
ii. The time of flight.
iii. The range.
iv. The maximum height.

Use [tex] g = 10 \, \text{m/s}^2 [/tex] and [tex] \sin^2 \theta = 2 \sin \theta \cos \theta [/tex].


Sagot :

To solve this problem, we'll use the principles of projectile motion. We'll break it down into steps to find the angle of projection, the time of flight, the range, and the maximum height.

### Step-by-Step Solution:

#### Given:
- Initial velocity (v) = 250 m/s
- Acceleration due to gravity (g) = 10 m/s²
- The maximum height (H) is one-third of the range (R)

#### i. Angle at which the bullet is fired (θ)

1. Equations of Projectile Motion:
- Maximum height (H):
[tex]\[ H = \frac{v^2 \sin^2(\theta)}{2g} \][/tex]
- Range (R):
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
- Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]

2. Substituting H = R/3 into equation:
[tex]\[ \frac{v^2 \sin^2(\theta)}{2g} = \frac{1}{3} \left( \frac{v^2 \sin(2\theta)}{g} \right) \][/tex]
Simplifying:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{\sin(2\theta)}{3} \][/tex]

3. Using the identity [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ \frac{\sin^2(\theta)}{2} = \frac{2 \sin(\theta) \cos(\theta)}{3} \][/tex]
[tex]\[ \frac{\sin(\theta)}{2 \cos(\theta)} = \frac{2}{3} \][/tex]
[tex]\[ \tan(\theta) = \frac{4}{3} \][/tex]

4. Finding θ:
[tex]\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 41.41^\circ \][/tex]

#### ii. Time of Flight (T)

1. Equation for Time of Flight:
[tex]\[ T = \frac{2v \sin(\theta)}{g} \][/tex]

2. Substitute values:
[tex]\[ T = \frac{2 \times 250 \times \sin(41.41^\circ)}{10} \][/tex]
[tex]\[ T \approx 33.07 \text{ seconds} \][/tex]

#### iii. Range (R)

1. Equation for Range:
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]

2. Substitute values:
[tex]\[ R = \frac{250^2 \cdot \sin(82.82^\circ)}{10} \][/tex]
[tex]\[ R \approx 6200.98 \text{ meters} \][/tex]

#### iv. Maximum Height (H)

1. Given relationship:
[tex]\[ H = \frac{R}{3} \][/tex]

2. Substitute range value:
[tex]\[ H = \frac{6200.98}{3} \][/tex]
[tex]\[ H \approx 2066.99 \text{ meters} \][/tex]

### Summary of Results:
1. Angle of Projection (θ): [tex]\(41.41^\circ\)[/tex]
2. Time of Flight (T): [tex]\(33.07\)[/tex] seconds
3. Range (R): [tex]\(6200.98\)[/tex] meters
4. Maximum Height (H): [tex]\(2066.99\)[/tex] meters
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.