A ball is projected horizontally. After 3s from projection its velocity becomes 1.25 times of the velocity of projection. Its velocity of projection is

Options

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Answer

To solve the problem step by step, we need to analyze the motion of the ball projected horizontally and how its velocity changes over time.

### Step 1: Understand the problem
A ball is projected horizontally, which means its initial vertical velocity (u_y) is 0 m/s. After 3 seconds, its resultant velocity becomes 1.25 times the horizontal component of its initial velocity (v_x).

### Step 2: Calculate the vertical velocity after 3 seconds
The vertical velocity (v_y) can be calculated using the formula:
\[ v_y = u_y + a_y \cdot t \]
Where:
- \( u_y = 0 \) m/s (initial vertical velocity)
- \( a_y = 9.8 \) m/s² (acceleration due to gravity)
- \( t = 3 \) s

Substituting the values:
\[ v_y = 0 + 9.8 \cdot 3 = 29.4 \text{ m/s} \]

### Step 3: Relate the resultant velocity to the horizontal velocity
The resultant velocity (v') after 3 seconds is given as:
\[ v' = 1.25 \cdot v_x \]
The resultant velocity can also be expressed in terms of its components:
\[ v' = \sqrt{v_x^2 + v_y^2} \]
Substituting for \( v_y \):
\[ v' = \sqrt{v_x^2 + (29.4)^2} \]

### Step 4: Set up the equation
Now we can set the two expressions for the resultant velocity equal to each other:
\[ 1.25 \cdot v_x = \sqrt{v_x^2 + (29.4)^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ (1.25 \cdot v_x)^2 = v_x^2 + (29.4)^2 \]
This simplifies to:
\[ 1.5625 \cdot v_x^2 = v_x^2 + 864.36 \]
Where \( 29.4^2 = 864.36 \).

### Step 6: Rearrange the equation
Rearranging gives:
\[ 1.5625 \cdot v_x^2 - v_x^2 = 864.36 \]
\[ 0.5625 \cdot v_x^2 = 864.36 \]

### Step 7: Solve for \( v_x^2 \)
Dividing both sides by 0.5625:
\[ v_x^2 = \frac{864.36}{0.5625} \]
Calculating this gives:
\[ v_x^2 \approx 1536 \]

### Step 8: Find \( v_x \)
Taking the square root:
\[ v_x \approx \sqrt{1536} \approx 39.2 \text{ m/s} \]

### Step 9: Round to the nearest option
Since the options provided are whole numbers, we round \( v_x \) to the nearest option, which is 40 m/s.

### Final Answer
The velocity of projection is approximately **40 m/s**.

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