A ball of mass 2 kg is projected horizontally with a velocity 20m/s from a building of height 15m. The speed with which body hits the ground is :

Options

A. 20 m/s
B. `10(7)^(1//2)m//s`
C. 25 m/s
D. 35 m/s

Answer

To find the speed with which the ball hits the ground, we need to consider both the horizontal and vertical components of the ball's motion. Here are the steps to solve the problem:

### Step 1: Identify the given data
- Mass of the ball (m) = 2 kg (not needed for speed calculation)
- Initial horizontal velocity (u_x) = 20 m/s
- Height of the building (h) = 15 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the time of flight
The time it takes for the ball to fall to the ground can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
Rearranging for \( t \):
\[ t = \sqrt{\frac{2h}{g}} \]
Substituting the values:
\[ t = \sqrt{\frac{2 \times 15 \, \text{m}}{10 \, \text{m/s}^2}} = \sqrt{3} \approx 1.73 \, \text{s} \]

### Step 3: Calculate the vertical velocity just before impact
The vertical velocity (\( v_y \)) just before hitting the ground can be calculated using the equation:
\[ v_y = g t \]
Substituting the values:
\[ v_y = 10 \, \text{m/s}^2 \times 1.73 \, \text{s} \approx 17.3 \, \text{m/s} \]

### Step 4: Calculate the resultant velocity just before impact
The resultant velocity (\( v \)) of the ball just before it hits the ground can be found using the Pythagorean theorem, since the horizontal and vertical components are perpendicular:
\[ v = \sqrt{u_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(20 \, \text{m/s})^2 + (17.3 \, \text{m/s})^2} \]
Calculating:
\[ v = \sqrt{400 + 299.29} \approx \sqrt{699.29} \approx 26.4 \, \text{m/s} \]

### Final Answer
The speed with which the body hits the ground is approximately **26.4 m/s**.

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