A body thrown vertically up from the ground passes the height of 102 m twice in an interval of 10 s. What was its initial velocity ?

Options

A. `52 ms^(-1)`
B. `61 ms^(-1)`
C. `45 ms^(-1)`
D. `26 ms^(-1)`

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration due to gravity.

### Step 1: Understand the motion of the body
The body is thrown vertically upwards and passes the height of 102 m twice in an interval of 10 seconds. This means it first reaches this height on its way up and then again on its way down.

### Step 2: Determine the time taken to reach the maximum height
Since the motion is symmetrical, the time taken to reach the maximum height (let's call it point B) from the ground (point A) is equal to the time taken to return to the same height from the maximum height (point C). Given that the total time from A to C is 10 seconds, the time taken to go from A to B (upwards) is:
\[ t_{AB} = \frac{10 \text{ s}}{2} = 5 \text{ s} \]

### Step 3: Use the first equation of motion
We can use the first equation of motion to find the final velocity (v) at the height of 102 m (point B):
\[ v = u - g t \]
Where:
- \( u \) is the initial velocity (which we need to find),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( t \) is the time taken to reach the height (5 s).

At the height of 102 m, the velocity will be:
\[ v = u - 10 \times 5 \]
\[ v = u - 50 \quad \text{(1)} \]

### Step 4: Use the second equation of motion
Now, we can use the second equation of motion to relate the initial velocity, final velocity, and displacement:
\[ v^2 = u^2 - 2g s \]
Where:
- \( s = 102 \, \text{m} \) (the height),
- \( g = 10 \, \text{m/s}^2 \).

Substituting the values, we have:
\[ v^2 = u^2 - 2 \times 10 \times 102 \]
\[ v^2 = u^2 - 2040 \quad \text{(2)} \]

### Step 5: Substitute equation (1) into equation (2)
From equation (1), we have \( v = u - 50 \). Squaring both sides gives:
\[ (u - 50)^2 = u^2 - 2040 \]
Expanding the left side:
\[ u^2 - 100u + 2500 = u^2 - 2040 \]

### Step 6: Simplify the equation
Cancelling \( u^2 \) from both sides:
\[ -100u + 2500 = -2040 \]
\[ -100u = -2040 - 2500 \]
\[ -100u = -4540 \]
\[ u = \frac{4540}{100} = 45.4 \, \text{m/s} \]

### Step 7: Final calculation
Thus, the initial velocity \( u \) is approximately:
\[ u = 45.4 \, \text{m/s} \]

### Conclusion
The initial velocity of the body thrown vertically upwards is approximately **45.4 m/s**.

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