. 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 problem, we need to use the concept of energy conservation in projectile motion. The maximum height \( h \) reached by an object in projectile motion can be derived using the principle of conservation of energy:

\[
\frac{1}{2} m v^2 = mgh
\]

Where:
- \( m \) is the mass of the object,
- \( v \) is the initial speed of the object,
- \( g \) is the acceleration due to gravity,
- \( h \) is the maximum height reached.

### Step 1: Analyze the first object
For the first object, we know:
- The maximum height is \( h_1 = 100 \, \text{m} \).
- The initial speed is \( v_1 \).

Using the energy conservation equation for the first object:
\[
\frac{1}{2} m v_1^2 = m g h_1
\]
\[
v_1^2 = 2 g h_1
\]
\[
v_1^2 = 2 g \times 100 = 200 g
\]
Thus, \( v_1^2 = 200g \).

### Step 2: Analyze the second object
For the second object:
- Its mass is \( \frac{2}{3} \) of the first object's mass.
- Its initial speed is \( v_2 = 2 v_1 \).

We need to find the maximum height \( h_2 \) it reaches. Again, using energy conservation:
\[
\frac{1}{2} \times \frac{2}{3} m \times v_2^2 = \frac{2}{3} m g h_2
\]
Substitute \( v_2 = 2 v_1 \):
\[
\frac{1}{2} \times \frac{2}{3} m \times (2 v_1)^2 = \frac{2}{3} m g h_2
\]
\[
\frac{1}{2} \times \frac{2}{3} m \times 4 v_1^2 = \frac{2}{3} m g h_2
\]
Substitute \( v_1^2 = 200 g \):
\[
\frac{1}{2} \times \frac{2}{3} m \times 4 \times 200 g = \frac{2}{3} m g h_2
\]
\[
\frac{1}{2} \times \frac{2}{3} m \times 800 g = \frac{2}{3} m g h_2
\]
The \( \frac{2}{3} m \) cancels out on both sides:
\[
\frac{1}{2} \times 800 g = g h_2
\]
\[
400 g = g h_2
\]
Thus, \( h_2 = 400 \, \text{m} \).

### Conclusion:
The second object reaches a maximum height of **400 meters**. Therefore, the correct answer is:

**(5) 400 m**

Would you like any more details or have any questions?

Here are 5 related questions to explore further:
1. What would happen to the height if the initial speed was tripled instead of doubled?
2. How does the mass of an object affect its maximum height if the speed is kept constant?
3. Can we find the time taken to reach the maximum height for each object?
4. What if the second object was thrown with the same speed as the first object?
5. How would air resistance affect the outcome in real-world scenarios?

**Tip:** Doubling the initial speed of an object quadruples its maximum height in ideal conditions (without air resistance).

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

Learn how to calculate the maximum height reached by two objects with different masses and speeds using energy conservation in projectile motion. This physics problem involves applying the energy formula to determine the height. Suitable for students in Grades 10-12.

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