A ball is thrown from ground such that it just crosses two poles of equal height kwpt 80 m apart. The maximum height attained by the ball is 80 m. When the ball passes the first pole, its velocity makes `45^(@)` with horizontal. The correct alternatives is/are :- `(g=10 m//s^(2))`

Options

A. Time interval between the two poles is 4s.
B. Height of the pole is 60 m.
C. Range of the ball is 160 m.
D. Angle of projection is `tan^(-1) (2)` with horizontal.

Answer

To solve the problem step by step, we will analyze the projectile motion of the ball thrown from the ground, considering the given conditions.

### Step 1: Understand the Problem
The ball is thrown from the ground and crosses two poles of equal height that are 80 m apart. The maximum height attained by the ball is 80 m, and when it passes the first pole, its velocity makes a 45-degree angle with the horizontal.

### Step 2: Determine the Velocity at the First Pole
When the ball passes the first pole, the angle of projection is 45 degrees. This means that the horizontal and vertical components of the velocity are equal.

Let the velocity at the first pole be \( v \). The horizontal and vertical components of the velocity can be expressed as:
- \( v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \)
- \( v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \)

### Step 3: Use the Range Formula
The range \( R \) for projectile motion is given by:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
Here, \( R = 80 \, \text{m} \), \( \theta = 45^\circ \), and \( g = 10 \, \text{m/s}^2 \).

Since \( \sin(90^\circ) = 1 \):
\[ 80 = \frac{v^2 \cdot 1}{10} \]
\[ v^2 = 800 \quad \Rightarrow \quad v = 20\sqrt{2} \, \text{m/s} \]

### Step 4: Calculate the Height at the First Pole
The height \( h' \) reached by the ball when it passes the first pole can be calculated using the formula:
\[ h' = \frac{v^2 \sin^2(\theta)}{2g} \]
Substituting \( v^2 = 800 \) and \( \theta = 45^\circ \):
\[ h' = \frac{800 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}{2 \cdot 10} = \frac{800 \cdot \frac{1}{2}}{20} = 20 \, \text{m} \]

### Step 5: Determine the Height of the Poles
The height of the poles is the maximum height minus the height at the first pole:
\[ \text{Height of the pole} = 80 \, \text{m} - h' = 80 \, \text{m} - 20 \, \text{m} = 60 \, \text{m} \]

### Step 6: Calculate the Time of Flight
The time of flight \( T \) from the first pole to the second pole can be calculated using:
\[ T = \frac{2v_y}{g} \]
Where \( v_y = v \sin(45^\circ) = \frac{20\sqrt{2}}{\sqrt{2}} = 20 \, \text{m/s} \):
\[ T = \frac{2 \cdot 20}{10} = 4 \, \text{s} \]

### Step 7: Calculate the Angle of Projection
The angle of projection \( \theta \) can be found using:
\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{40}{20} = 2 \]
Thus, \( \theta = \tan^{-1}(2) \).

### Step 8: Calculate the Complete Range
The complete range \( R \) can be calculated using:
\[ R = \frac{2 v_x v_y}{g} \]
Substituting \( v_x = 20 \, \text{m/s} \) and \( v_y = 40 \, \text{m/s} \):
\[ R = \frac{2 \cdot 20 \cdot 40}{10} = 160 \, \text{m} \]

### Summary of Results
1. Height of the poles: 60 m
2. Time of flight: 4 s
3. Angle of projection: \( \tan^{-1}(2) \)
4. Complete range: 160 m