A body, thrown upwards with some velocity, reaches the maximum height of 20 m. Another body with double the mass thrown up, with double intial velocity will each a maximum height of

Options

A. 40 m
B. 80 m
C. 120 m
D. 160 m

Explanation

To solve the problem, we need to determine the maximum height reached by the second body, which is thrown upwards with double the mass and double the initial velocity of the first body.

### Step-by-Step Solution:

1. **Identify the known values:**
- The maximum height \( h_1 \) reached by the first body is 20 m.
- The initial velocity of the first body is \( u_1 \).
- The second body has double the mass and double the initial velocity, so its initial velocity \( u_2 = 2u_1 \).

2. **Use the kinematic equation for maximum height:**
The maximum height \( h \) reached by an object thrown upwards can be calculated using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0 m/s at maximum height),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \(-g\), where \( g \approx 9.8 \, \text{m/s}^2 \)),
- \( s \) is the maximum height \( h \).

3. **Apply the equation for the first body:**
For the first body, at maximum height:
\[ 0 = u_1^2 - 2gh_1 \]
Rearranging gives:
\[ u_1^2 = 2gh_1 \]
Substituting \( h_1 = 20 \, \text{m} \):
\[ u_1^2 = 2g \cdot 20 \]

4. **Apply the equation for the second body:**
For the second body, we have:
\[ 0 = u_2^2 - 2gh_2 \]
Substituting \( u_2 = 2u_1 \):
\[ 0 = (2u_1)^2 - 2gh_2 \]
This simplifies to:
\[ 0 = 4u_1^2 - 2gh_2 \]
Rearranging gives:
\[ 4u_1^2 = 2gh_2 \]
Now, substituting \( u_1^2 \) from the first body's equation:
\[ 4(2gh_1) = 2gh_2 \]

5. **Solve for \( h_2 \):**
\[ 8gh_1 = 2gh_2 \]
Dividing both sides by \( 2g \):
\[ 4h_1 = h_2 \]
Substituting \( h_1 = 20 \, \text{m} \):
\[ h_2 = 4 \cdot 20 = 80 \, \text{m} \]

### Final Answer:
The maximum height reached by the second body is **80 meters**.