A body falling a vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10 s. The height h is `(g=10m//s^(2))`

Options

A. 22 m
B. 10.2 m
C. 11.2 m
D. 15 m/s

Answer

To solve the problem, we need to determine the height \( h \) at which a body, projected vertically upwards with an initial velocity of 52 m/s, passes twice at an interval of 10 seconds. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The body is projected upwards with an initial velocity \( u = 52 \, \text{m/s} \). It will rise to a maximum height and then fall back down. The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards.

2. **Time Interval**:
The body passes the height \( h \) twice in a total time interval of 10 seconds. This means it takes 5 seconds to reach the height \( h \) on the way up and another 5 seconds to return to the same height on the way down.

3. **Finding the Velocity at Maximum Height**:
At the maximum height, the final velocity \( v = 0 \). We can use the first equation of motion:
\[ v = u - gt \]
Substituting the values:
\[ 0 = 52 - 10 \cdot 5 \]
Simplifying gives:
\[ 0 = 52 - 50 \implies u = 52 \, \text{m/s} \]
This confirms that the body reaches the maximum height after 5 seconds.

4. **Finding the Maximum Height**:
We can use the second equation of motion to find the maximum height \( h_{max} \):
\[ s = ut - \frac{1}{2}gt^2 \]
Substituting \( u = 52 \, \text{m/s} \), \( g = 10 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
\[ s = 52 \cdot 5 - \frac{1}{2} \cdot 10 \cdot (5^2) \]
Simplifying gives:
\[ s = 260 - \frac{1}{2} \cdot 10 \cdot 25 = 260 - 125 = 135 \, \text{m} \]

5. **Finding the Height \( h \)**:
The body passes the height \( h \) twice: once while going up and once while coming down. The time taken to reach height \( h \) from the ground is 5 seconds. We can use the equation of motion again to find \( h \):
\[ v^2 = u^2 - 2gh \]
At the point where it is at height \( h \) on the way up, we can find the velocity \( v \) at that height using:
\[ v = u - gt \]
Substituting \( t = 5 \):
\[ v = 52 - 10 \cdot 5 = 2 \, \text{m/s} \]
Now substituting \( u = 52 \, \text{m/s} \), \( v = 2 \, \text{m/s} \), and \( g = 10 \, \text{m/s}^2 \) into the equation:
\[ 2^2 = 52^2 - 2 \cdot 10 \cdot h \]
Simplifying gives:
\[ 4 = 2704 - 20h \]
Rearranging gives:
\[ 20h = 2704 - 4 = 2700 \]
Therefore:
\[ h = \frac{2700}{20} = 135 \, \text{m} \]

### Final Answer:
The height \( h \) is \( 135 \, \text{m} \).