To solve Question 6, let's break it into parts:

Question Recap:

A bullet is fired at an angle of $45^\circ$ with the horizontal with a velocity of $275 \, \text{m/s}$.
Find:

1. The maximum height the bullet will rise above the ground.
2. The horizontal range of the bullet.
   Take $g = 9.8 \, \text{m/s}^2$.

Part 1: Maximum Height ($H_\text{max}$)

The formula for maximum height in projectile motion is:

$H_\text{max} = \frac{u_y^2}{2g}$

Where:

• $u_y = u \sin\theta$ is the vertical component of the velocity.
• $g$ is the acceleration due to gravity.

Step 1: Find $u_y$

$u_y = u \sin\theta = 275 \sin 45^\circ$

From trigonometry, $\sin 45^\circ = \frac{\sqrt{2}}{2}$: $u_y = 275 \cdot \frac{\sqrt{2}}{2} = 194.51 \, \text{m/s} \, (\text{approx.})$

Step 2: Substitute $u_y$ into $H_\text{max}$

$H_\text{max} = \frac{u_y^2}{2g} = \frac{(194.51)^2}{2 \cdot 9.8}$

$H_\text{max} = \frac{37830.89}{19.6} = 1930.15 \, \text{m}$

Maximum Height:
$H_\text{max} \approx 1930.15 \, \text{m}$

Part 2: Horizontal Range ($R$)

The formula for horizontal range in projectile motion is:

$R = \frac{u^2 \sin 2\theta}{g}$

Where:

• $u$ is the initial velocity,
• $\sin 2\theta = \sin(2 \cdot 45^\circ) = \sin 90^\circ = 1$,
• $g$ is the acceleration due to gravity.

Step 1: Substitute into the formula:

$R = \frac{u^2 \cdot 1}{g} = \frac{(275)^2}{9.8}$

$R = \frac{75625}{9.8} = 7711.73 \, \text{m}$

Horizontal Range:
$R \approx 7711.73 \, \text{m}$

Final Answer:

1. Maximum Height: 1930.15 m
2. Horizontal Range: 7711.73 m

Do you want me to clarify any part of the solution or help with another question?

Related Questions:

1. What is the total time of flight for the bullet in this scenario?
2. Derive the formula for maximum height in projectile motion.
3. If the bullet were fired at a different angle, how would the range and height change?
4. How does air resistance affect the horizontal range and maximum height?
5. What is the velocity of the bullet at the maximum height?

Tip:

For maximum range in projectile motion, the launch angle should be $45^\circ$.