At time `t=0`, a small ball is projected from point A with a velocity of `60m//s` at `60^@` angle with horizontal. Neglect atmospheric resistance and determine the two times `t_1 and t_2` when the velocity of the ball makes an angle of `45^@` with horizontal x-axis.

Explanation

To solve the problem, we need to analyze the projectile motion of the ball and find the times \( t_1 \) and \( t_2 \) when the velocity of the ball makes an angle of \( 45^\circ \) with the horizontal.

### Step 1: Determine the initial velocity components
The initial velocity \( u \) of the ball is given as \( 60 \, \text{m/s} \) at an angle of \( 60^\circ \).

- The horizontal component of the initial velocity \( u_x \) is given by:
\[ u_x = u \cos(60^\circ) = 60 \cos(60^\circ) = 60 \times \frac{1}{2} = 30 \, \text{m/s} \]

- The vertical component of the initial velocity \( u_y \) is given by:
\[ u_y = u \sin(60^\circ) = 60 \sin(60^\circ) = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{m/s} \]

### Step 2: Set up the equations for the vertical velocity
At the times \( t_1 \) and \( t_2 \), the vertical component of the velocity \( v_y \) must equal the horizontal component \( u_x \) for the angle to be \( 45^\circ \):
\[ v_y = u_x = 30 \, \text{m/s} \]

The vertical velocity \( v_y \) at time \( t \) is given by:
\[ v_y = u_y - g t \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Solve for \( t_1 \)
For the first instance when the ball is at \( 45^\circ \):
\[ 30 = 30\sqrt{3} - 10 t_1 \]
Rearranging gives:
\[ 10 t_1 = 30\sqrt{3} - 30 \]
\[ t_1 = \frac{30(\sqrt{3} - 1)}{10} = 3(\sqrt{3} - 1) \approx 2.86 \, \text{s} \]

### Step 4: Solve for \( t_2 \)
For the second instance when the ball is at \( 45^\circ \):
At this point, the vertical velocity will be negative:
\[ -30 = 30\sqrt{3} - 10 t_2 \]
Rearranging gives:
\[ 10 t_2 = 30\sqrt{3} + 30 \]
\[ t_2 = \frac{30(\sqrt{3} + 1)}{10} = 3(\sqrt{3} + 1) \approx 8.20 \, \text{s} \]

### Final Answer
The two times when the velocity of the ball makes an angle of \( 45^\circ \) with the horizontal are:
\[ t_1 \approx 2.86 \, \text{s} \quad \text{and} \quad t_2 \approx 8.20 \, \text{s} \]