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 step-by-step, we need to analyze the motion of the body projected upwards with an initial velocity of 52 m/s and determine the height \( h \) at which it passes twice in an interval of 10 seconds.

### Step 1: Understand the Motion
The body is projected upwards with an initial velocity \( u = 52 \, \text{m/s} \). It will rise to a certain height, come to a stop momentarily, and then fall back down, passing the same height \( h \) twice during its motion.

### Step 2: Use the Kinematic Equation
We can use the kinematic equation for the velocity of the body at height \( h \):
\[ v^2 = u^2 - 2gh \]
where:
- \( v \) is the velocity at height \( h \),
- \( u = 52 \, \text{m/s} \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h \) is the height we want to find.

### Step 3: Analyze the Time of Flight
The body takes 10 seconds to go up to the height \( h \) and then 10 seconds to come back down. Therefore, the total time taken to reach the maximum height and return to height \( h \) is 20 seconds.

### Step 4: Calculate the Displacement
Using the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
For the upward journey to height \( h \):
\[ h = ut - \frac{1}{2} g t^2 \]
For the upward journey, the time \( t \) to reach the maximum height is \( 10 \, \text{s} \):
\[ h = 52 \cdot 10 - \frac{1}{2} \cdot 10 \cdot (10)^2 \]
\[ h = 520 - 500 = 20 \, \text{m} \]

### Step 5: Calculate the Velocity at Height \( h \)
Now, we can find the velocity \( v \) when the body reaches height \( h \):
\[ v^2 = 52^2 - 2 \cdot 10 \cdot h \]
Substituting \( h = 20 \):
\[ v^2 = 52^2 - 2 \cdot 10 \cdot 20 \]
\[ v^2 = 2704 - 400 = 2304 \]
\[ v = \sqrt{2304} = 48 \, \text{m/s} \]

### Step 6: Solve for Height \( h \)
Now, we can set up the equation for the total displacement when the body returns to height \( h \):
\[ 0 = 52 \cdot 10 + \frac{1}{2} (-10) \cdot (10)^2 \]
This simplifies to:
\[ 0 = 520 - 500 \]
This confirms that the body returns to the same height \( h \) after 10 seconds.

### Final Answer
The height \( h \) is \( 20 \, \text{m} \).