A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The speed with which it hits the ground in `ms^(-1)` is `(g=9.8ms^(2))`

Options

A. 9.8
B. 99
C. 4.9
D. 49

Explanation

To solve the problem of the hiker throwing a stone horizontally from a cliff, we will follow these steps:

### Step 1: Determine the time taken to hit the ground
We know the height of the cliff (h) is 490 m, and the initial vertical velocity (u_y) is 0 m/s (since the stone is thrown horizontally). We can use the following kinematic equation for vertical motion:

\[ h = u_y t + \frac{1}{2} g t^2 \]

Substituting the values:

\[ 490 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

This simplifies to:

\[ 490 = 4.9 t^2 \]

Now, solving for \(t^2\):

\[ t^2 = \frac{490}{4.9} = 100 \]

Taking the square root:

\[ t = 10 \text{ seconds} \]

### Step 2: Calculate the horizontal velocity
The horizontal velocity (v_x) remains constant because there is no horizontal acceleration (neglecting air resistance). Thus:

\[ v_x = 15 \text{ m/s} \]

### Step 3: Calculate the vertical velocity just before hitting the ground
Using the vertical motion equation, we can find the final vertical velocity (v_y) just before hitting the ground:

\[ v_y = u_y + g t \]

Substituting the values:

\[ v_y = 0 + 9.8 \cdot 10 = 98 \text{ m/s} \]

### Step 4: Calculate the resultant velocity when the stone hits the ground
The resultant velocity (v) can be found using the Pythagorean theorem, since the horizontal and vertical components are perpendicular to each other:

\[ v = \sqrt{v_x^2 + v_y^2} \]

Substituting the values:

\[ v = \sqrt{(15)^2 + (98)^2} = \sqrt{225 + 9604} = \sqrt{9829} \]

Calculating the square root:

\[ v \approx 99.14 \text{ m/s} \]

### Final Answer
Thus, the speed with which the stone hits the ground is approximately **99.14 m/s**.

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