A stone is thrown horizontally with velocity u. The velocity of the stone 0.5 s later is 3u/2. The value of u is

Options

A. `2.2 m s^(-1)`
B. `3.3 m s^(-1)`
C. `4.4 m s^(-1)`
D. `1.1 m s^(-1)`

Answer

To solve the problem, we will break it down into steps:

### Step 1: Understand the problem
A stone is thrown horizontally with an initial velocity \( u \). After \( 0.5 \) seconds, its velocity becomes \( \frac{3u}{2} \). We need to find the value of \( u \).

### Step 2: Identify the components of velocity
When the stone is thrown horizontally, its velocity has two components:
- Horizontal component \( V_x = u \) (remains constant since there is no horizontal acceleration)
- Vertical component \( V_y \) (changes due to gravitational acceleration)

### Step 3: Calculate the vertical component of velocity after \( 0.5 \) seconds
The vertical component of velocity can be calculated using the formula:
\[ V_y = V_{y0} + a_y \cdot t \]
where:
- \( V_{y0} = 0 \) (initial vertical velocity)
- \( a_y = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 0.5 \, \text{s} \)

Substituting the values:
\[ V_y = 0 + 10 \cdot 0.5 = 5 \, \text{m/s} \]

### Step 4: Use the Pythagorean theorem to relate the velocities
The resultant velocity \( V \) after \( 0.5 \) seconds is given by:
\[ V = \sqrt{V_x^2 + V_y^2} \]
We know that \( V = \frac{3u}{2} \), \( V_x = u \), and \( V_y = 5 \, \text{m/s} \). Therefore:
\[ \frac{3u}{2} = \sqrt{u^2 + 5^2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{3u}{2}\right)^2 = u^2 + 25 \]
This simplifies to:
\[ \frac{9u^2}{4} = u^2 + 25 \]

### Step 6: Rearrange the equation
Rearranging the equation:
\[ \frac{9u^2}{4} - u^2 = 25 \]
To combine the terms, express \( u^2 \) with a common denominator:
\[ \frac{9u^2}{4} - \frac{4u^2}{4} = 25 \]
This simplifies to:
\[ \frac{5u^2}{4} = 25 \]

### Step 7: Solve for \( u^2 \)
Multiplying both sides by \( 4 \):
\[ 5u^2 = 100 \]
Dividing by \( 5 \):
\[ u^2 = 20 \]

### Step 8: Solve for \( u \)
Taking the square root of both sides:
\[ u = \sqrt{20} = 4.47 \, \text{m/s} \, (\text{approximately}) \]

### Final Answer
The value of \( u \) is approximately \( 4.47 \, \text{m/s} \).

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