A boy whirls a stone of small mass in a horizontal circle of radius `1.5m` and at height `2.9m` above level ground. The string breacks and the stone flies off horizontally and strikes the ground after travelling a horizontal distance of `10m` . What is the magnitude of the centripetal acceleration of the stone while in circular motion ?

Options

A. `33ms^(-2)`
B. `63ms^(-2)`
C. `93ms^(-2)`
D. `113ms^(-2)`

Explanation

To find the magnitude of the centripetal acceleration of the stone while in circular motion, we will follow these steps:

### Step 1: Identify the known values
- Radius of the circular motion, \( R = 1.5 \, \text{m} \)
- Height above ground, \( h = 2.9 \, \text{m} \)
- Horizontal distance traveled after the string breaks, \( d = 10 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the time of flight
When the stone flies off horizontally, it falls from a height of \( h = 2.9 \, \text{m} \). The time \( t \) it takes to hit the ground can be calculated using the formula for free fall:

\[ t = \sqrt{\frac{2h}{g}} \]

Substituting the known values:

\[ t = \sqrt{\frac{2 \times 2.9}{9.8}} = \sqrt{\frac{5.8}{9.8}} \approx \sqrt{0.59183673} \approx 0.77 \, \text{s} \]

### Step 3: Calculate the horizontal velocity
The horizontal velocity \( v \) of the stone when it leaves the circular motion can be calculated using the formula:

\[ v = \frac{d}{t} \]

Substituting the known values:

\[ v = \frac{10 \, \text{m}}{0.77 \, \text{s}} \approx 12.99 \, \text{m/s} \approx 13 \, \text{m/s} \]

### Step 4: Calculate the centripetal acceleration
The centripetal acceleration \( a_c \) is given by the formula:

\[ a_c = \frac{v^2}{R} \]

Substituting the values of \( v \) and \( R \):

\[ a_c = \frac{(13 \, \text{m/s})^2}{1.5 \, \text{m}} = \frac{169}{1.5} \approx 112.67 \, \text{m/s}^2 \approx 113 \, \text{m/s}^2 \]

### Conclusion
The magnitude of the centripetal acceleration of the stone while in circular motion is approximately \( 113 \, \text{m/s}^2 \).

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