Given that the two objects are projected with the same velocity but different angles, and they cover the same range, we are asked to find the maximum height of the second object when the first object's maximum height is given as $h$.

Key concepts:

1. Range of a projectile is given by the formula: $R = \frac{v_0^2 \sin(2\theta)}{g}$ where:

   • $v_0$ is the initial velocity,
   • $\theta$ is the launch angle,
   • $g$ is the acceleration due to gravity.

2. Maximum height of a projectile is given by: $H = \frac{v_0^2 \sin^2(\theta)}{2g}$ where $\theta$ is the angle of projection.

Since both projectiles cover the same range, we know that: $R_1 = R_2$ This implies: $\frac{v_0^2 \sin(2\theta_1)}{g} = \frac{v_0^2 \sin(2\theta_2)}{g}$ Simplifying, we get: $\sin(2\theta_1) = \sin(2\theta_2)$ This means that $2\theta_1 = \pi - 2\theta_2$ (since $\sin(\alpha) = \sin(\pi - \alpha)$), so: $\theta_2 = \frac{\pi}{2} - \theta_1$

Given:

• $\theta_1 = \frac{\pi}{3}$
• The maximum height for the first projectile is $H_1 = h$.

For the first projectile:

The maximum height is: $h = \frac{v_0^2 \sin^2(\theta_1)}{2g} = \frac{v_0^2 \sin^2\left(\frac{\pi}{3}\right)}{2g}$ Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, we get: $h = \frac{v_0^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{v_0^2 \cdot \frac{3}{4}}{2g} = \frac{3v_0^2}{8g}$

For the second projectile:

The angle $\theta_2$ is $\frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.

The maximum height for the second projectile is: $H_2 = \frac{v_0^2 \sin^2(\theta_2)}{2g} = \frac{v_0^2 \sin^2\left(\frac{\pi}{6}\right)}{2g}$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we get: $H_2 = \frac{v_0^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{v_0^2 \cdot \frac{1}{4}}{2g} = \frac{v_0^2}{8g}$

Relation between $h$ and $H_2$:

From the equation for $h$: $h = \frac{3v_0^2}{8g}$ we can now express $H_2$ in terms of $h$: $H_2 = \frac{v_0^2}{8g} = \frac{h}{3}$

Final Answer:

The maximum height of the second projectile is $\frac{h}{3}$.

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Would you like more details on any part of this solution or have any questions?

Here are some additional questions for further exploration:

1. How does the launch angle affect the range of a projectile?
2. How would the maximum height change if the initial velocity was doubled?
3. What is the impact of air resistance on the range and height of a projectile?
4. How do we derive the formula for the range of a projectile?
5. How can you calculate the time of flight for a projectile?

Tip: When solving projectile motion problems, always start by analyzing the given quantities (velocity, angle, range) and use the basic kinematic equations to relate them.