A ball is thrown with speed `40 m//s` at an angle `30^(@)` with horizontally from the top of a tower of height `60 m`. Choose the correct option

Options

A. the vertical component of velocity first decreases to zero and then increases
B. the ball reaches the ground after `6 s`
C. if the ball strikes the ground at maximum horizontal distance from the tower for this the angle of projection should be less than `45^(@)`
D. all option are correct

Explanation

To solve the problem step by step, we need to analyze the motion of the ball thrown from the tower. Here’s how we can approach it:

### Step 1: Identify the initial parameters
- Initial speed of the ball, \( u = 40 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Height of the tower, \( h = 60 \, \text{m} \)

### Step 2: Break down the initial velocity into components
The initial velocity can be resolved into horizontal and vertical components:
- \( u_x = u \cos \theta = 40 \cos 30^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{m/s} \)
- \( u_y = u \sin \theta = 40 \sin 30^\circ = 40 \times \frac{1}{2} = 20 \, \text{m/s} \)

### Step 3: Use the vertical motion equation to find the time of flight
The vertical motion can be described by the equation:
\[ y = u_y t - \frac{1}{2} g t^2 \]
where \( y \) is the vertical displacement (which will be -60 m since it falls down), \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity), and \( t \) is the time of flight.

Substituting the values:
\[ -60 = 20t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ -60 = 20t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 20t - 60 = 0 \]
Dividing through by 5:
\[ t^2 - 4t - 12 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4, c = -12 \):
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ t = \frac{4 \pm \sqrt{16 + 48}}{2} \]
\[ t = \frac{4 \pm \sqrt{64}}{2} \]
\[ t = \frac{4 \pm 8}{2} \]
This gives us two possible solutions:
\[ t = \frac{12}{2} = 6 \quad \text{and} \quad t = \frac{-4}{2} = -2 \, (\text{not valid}) \]
Thus, \( t = 6 \, \text{s} \).

### Step 5: Verify the options
1. **Vertical component of velocity first decreases to zero then increases**: This is true because the ball goes up first, slows down, stops, and then falls down.
2. **The ball reaches the ground after 6 seconds**: This is confirmed as we calculated \( t = 6 \, \text{s} \).
3. **If the ball strikes the ground at maximum horizontal distance from the tower, the angle of projection should be less than 45 degrees**: This is also true since for maximum range, the angle should be less than 45 degrees when launched from a height.

### Conclusion
All options provided in the question are correct.

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