A bal is thrown vertically upwards from the ground. It crosses a point at the height of 25 m twice at an interval of 4 secs. The ball was thrown with the velocity of

Options

A. 20 m/s
B. 25 m/s
C. 30 m/s
D. 35 m/s

Explanation

To solve the problem of finding the initial velocity of a ball thrown vertically upwards that crosses a height of 25 m twice at an interval of 4 seconds, we can follow these steps:

### Step 1: Understand the Motion of the Ball
The ball is thrown upwards and reaches a maximum height before falling back down. It crosses the height of 25 m twice: once on the way up and once on the way down. The time interval between these two crossings is given as 4 seconds.

### Step 2: Determine the Time to Reach Maximum Height
Since the ball takes 4 seconds to cross the height of 25 m twice, it means that it takes 2 seconds to go from the height of 25 m to the maximum height (point C) and another 2 seconds to come back down from the maximum height to the height of 25 m again.

### Step 3: Use the Kinematic Equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
\[ v = u + at \]
Where:
- \( v \) is the final velocity (0 m/s at the maximum height),
- \( u \) is the initial velocity (which we need to find),
- \( a \) is the acceleration due to gravity (-10 m/s², negative because it acts downwards),
- \( t \) is the time taken to reach the maximum height (2 seconds).

### Step 4: Calculate the Initial Velocity at Point B
At the maximum height, the final velocity \( v = 0 \). Rearranging the equation gives:
\[ 0 = u - 10 \times 2 \]
\[ u = 20 \, \text{m/s} \]
So, the velocity at point B (just before reaching the height of 25 m) is 20 m/s.

### Step 5: Use Another Kinematic Equation to Find Initial Velocity
Now, we can use another kinematic equation to find the initial velocity of the ball when it was thrown:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity at the height of 25 m (which we found to be 20 m/s),
- \( u \) is the initial velocity we want to find,
- \( a \) is the acceleration due to gravity (-10 m/s²),
- \( s \) is the height (25 m).

### Step 6: Substitute Values into the Equation
Substituting the known values into the equation:
\[ (20)^2 = u^2 - 2 \times 10 \times 25 \]
\[ 400 = u^2 - 500 \]
\[ u^2 = 400 + 500 \]
\[ u^2 = 900 \]

### Step 7: Solve for Initial Velocity
Taking the square root of both sides gives:
\[ u = \sqrt{900} \]
\[ u = 30 \, \text{m/s} \]

### Conclusion
Thus, the initial velocity with which the ball was thrown is **30 m/s**.

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