A ball is rolled off the edge of a horizontal table at a speed of `4 m//"second"`. It hits the ground after 0.4 second . Which statement given below is true

Options

A. It hits the ground at a horizontal distance 1.6 m from the edge of the table
B. The speed with which it hits the ground is 4.0 m/second
C. Height of the table is `0.8 m `
D. It hits the ground at an angle of `60^(@)` to the horizontal

Answer

To solve the problem step by step, we need to analyze the motion of the ball after it rolls off the edge of the table.

### Step 1: Determine the vertical distance (height of the table)
The ball is rolled off the table horizontally with an initial horizontal velocity of \( u_x = 4 \, \text{m/s} \). The time taken to hit the ground is \( t = 0.4 \, \text{s} \).

In the vertical direction, the initial vertical velocity \( u_y = 0 \) (since it is rolled off horizontally). The only force acting on the ball in the vertical direction is gravity, which causes a downward acceleration \( a = g = 10 \, \text{m/s}^2 \).

Using the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ S = 0 \cdot 0.4 + \frac{1}{2} \cdot 10 \cdot (0.4)^2 \]
\[ S = 0 + \frac{1}{2} \cdot 10 \cdot 0.16 \]
\[ S = 5 \cdot 0.16 = 0.8 \, \text{m} \]
Thus, the height of the table is \( 0.8 \, \text{m} \).

### Step 2: Calculate the horizontal distance traveled
In the horizontal direction, the horizontal velocity remains constant since there is no acceleration. The horizontal distance \( S_x \) can be calculated as:
\[ S_x = u_x \cdot t \]
Substituting the values:
\[ S_x = 4 \, \text{m/s} \cdot 0.4 \, \text{s} = 1.6 \, \text{m} \]
Thus, the ball travels a horizontal distance of \( 1.6 \, \text{m} \) before hitting the ground.

### Step 3: Determine the final vertical velocity
To find the final vertical velocity \( v_y \) just before hitting the ground, we can use the first equation of motion:
\[ v_y^2 = u_y^2 + 2aS \]
Substituting the values:
\[ v_y^2 = 0 + 2 \cdot 10 \cdot 0.8 \]
\[ v_y^2 = 16 \implies v_y = \sqrt{16} = 4 \, \text{m/s} \]

### Step 4: Calculate the resultant velocity just before hitting the ground
The resultant velocity \( v \) just before hitting the ground can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Where \( v_x = 4 \, \text{m/s} \) (horizontal velocity) and \( v_y = 4 \, \text{m/s} \) (vertical velocity):
\[ v = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \, \text{m/s} \approx 5.66 \, \text{m/s} \]

### Step 5: Determine the angle of impact
The angle \( \theta \) with respect to the horizontal can be found using:
\[ \tan(\theta) = \frac{v_y}{v_x} \]
Substituting the values:
\[ \tan(\theta) = \frac{4}{4} = 1 \implies \theta = 45^\circ \]

### Conclusion
Based on the calculations:
1. The height of the table is \( 0.8 \, \text{m} \).
2. The horizontal distance traveled is \( 1.6 \, \text{m} \).
3. The speed at which it hits the ground is approximately \( 5.66 \, \text{m/s} \).
4. The angle of impact is \( 45^\circ \).

### True Statements
- The ball hits the ground at a horizontal distance of \( 1.6 \, \text{m} \) from the edge of the table.
- The height of the table is \( 0.8 \, \text{m} \).