. An object projected upwards reaches a maximum height of 100 m. When an object with a mass two-thirds that of the first object is thrown upwards with a speed twice the projectile speed of the first object, the maximum height it reaches is,

(1) 25 m

(2) 50 m

(3) 100 m

(4) 200 m

(5) 400 m

To solve this, we use the concept of energy conservation and the equations for the motion of projectiles.

Let’s break down the problem into two parts:

1. **For the first object**:
   - Maximum height \( H_1 = 100 \, \text{m} \)
   - At the maximum height, the object's kinetic energy has been fully converted into potential energy. The relationship between initial kinetic energy and potential energy at the maximum height is:
     \[
     \frac{1}{2} m v_1^2 = m g H_1
     \]
     where:
     - \( m \) is the mass of the first object,
     - \( v_1 \) is the initial speed of the first object,
     - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)),
     - \( H_1 \) is the maximum height reached by the first object.

     Simplifying this equation:
     \[
     v_1^2 = 2 g H_1
     \]

2. **For the second object**:
   - The mass of the second object is \( \frac{2}{3} \) that of the first, i.e., \( m_2 = \frac{2}{3} m \).
   - The initial speed of the second object is \( 2 v_1 \).
   - The maximum height reached by the second object, \( H_2 \), can be determined by equating the initial kinetic energy to the potential energy at maximum height:
     \[
     \frac{1}{2} m_2 (2v_1)^2 = m_2 g H_2
     \]
     Substituting \( m_2 = \frac{2}{3} m \) and simplifying:
     \[
     \frac{1}{2} \times \frac{2}{3} m \times 4 v_1^2 = \frac{2}{3} m g H_2
     \]
     \[
     \frac{8}{3} \times \frac{1}{2} m v_1^2 = \frac{2}{3} m g H_2
     \]
     Using the relation \( v_1^2 = 2 g H_1 \):
     \[
     \frac{8}{3} \times g H_1 = \frac{2}{3} g H_2
     \]
     Canceling out common terms:
     \[
     8 H_1 = 2 H_2
     \]
     \[
     H_2 = 4 H_1
     \]
     Since \( H_1 = 100 \, \text{m} \), we have:
     \[
     H_2 = 4 \times 100 = 400 \, \text{m}
     \]

Thus, the maximum height reached by the second object is **400 m**, corresponding to option (5).

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Do you want more details or have any questions?

Here are 5 related questions:

1. How does the mass of an object affect its maximum height in projectile motion?
2. How would the problem change if air resistance were considered?
3. What happens to the total mechanical energy of the object during its upward and downward journey?
4. How do different initial angles of projection affect the height and range of the projectile?
5. Can you derive the equation of motion for a projectile in two dimensions?

**Tip**: Always check if kinetic and potential energy are fully conserved when dealing with ideal projectile motion.

An object projected upwards reaches a maximum height of 100 m. When an object with a mass two-thirds that of the first object is thrown upwards with a speed twice the projectile speed of the first object, the maximum height it reaches is:

(1) 25 m
(2) 50 m
(3) 100 m
(4) 200 m
(5) 400 m

Solve a projectile motion problem where an object reaches a maximum height of 100 m. Learn how energy conservation principles help calculate the maximum height of a second object with a different mass and velocity. Detailed steps for students in Grades 11-12.

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