To calculate the maximum height and range of a projectile, we will use the following physics equations based on projectile motion.

Given dataTo calculate the maximum height and range of a projectile, we will use the following physics equations based on projectile motion.

Given

• Initial speed, $v_0 = 40 \, \text{m/s}$
• Angle of projection, $\theta = 60^\circ$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

1. Maximum Height

The formula for the maximum height $H$ is:

$H = \frac{(v_0 \sin \theta)^2}{2g}$

Let's break it down:

• Vertical component of velocity: $v_{0y} = v_0 \sin \theta$

Plugging in the values:

$v_{0y} = 40 \sin 60^\circ = 40 \times \frac{\sqrt{3}}{2} = 40 \times 0.866 = 34.64 \, \text{m/s}$

Now, applying the formula for maximum height:

$H = \frac{(34.64)^2}{2 \times 9.8} = \frac{1199.06}{19.6} = 61.17 \, \text{m}$

2. Range

The formula for the range $R$ of a projectile is:

$R = \frac{v_0^2 \sin 2\theta}{g}$

We need $\sin 2\theta$: $\sin 2\theta = \sin (2 \times 60^\circ) = \sin 120^\circ = \sin 60^\circ = 0.866$

Now, using the range formula:

$R = \frac{40^2 \times 0.866}{9.8} = \frac{1600 \times 0.866}{9.8} = \frac{1385.6}{9.8} = 141.39 \, \text{m}$

Final Answers:

• Maximum height: $61.17 \, \text{m}$
• Range: $141.39 \, \text{m}$

Would you like more details or clarifications on any part of the solution?

Here are 5 related questions for practice:

1. What is the time of flight for this projectile?
2. How does changing the angle affect the maximum height and range?
3. What would be the maximum height if the launch speed was 50 m/s instead?
4. How would air resistance affect the range of the projectile?
5. What is the horizontal velocity of the projectile throughout its flight?

Tip: In projectile motion, the horizontal velocity remains constant, but the vertical velocity changes due to gravity.