A ball is projected at an angle `60^(@)` with the horizontal with speed 30 m/s. What will be the speed of the ball when it makes an angle `45^(@)` with the horizontal ?

Options

A. 30 m/s
B. `15sqrt2m//s`
C. `(15)/(sqrt2)m//s`
D. `30sqrt2m//s`

Answer

To solve the problem, we will analyze the projectile motion of the ball and use the concepts of horizontal and vertical components of velocity.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The initial speed \( u = 30 \, \text{m/s} \)
- The angle of projection \( \theta = 60^\circ \)

2. **Calculate the Horizontal Component of Initial Velocity:**
\[ u_x = u \cos \theta = 30 \cos 60^\circ \]
Since \( \cos 60^\circ = \frac{1}{2} \):
\[ u_x = 30 \times \frac{1}{2} = 15 \, \text{m/s} \]

3. **Determine the Horizontal Component of Velocity at \( 45^\circ \):**
- When the ball makes an angle \( \phi = 45^\circ \) with the horizontal, the horizontal component of the velocity \( v_x \) remains the same (as there is no horizontal acceleration):
\[ v_x = u_x = 15 \, \text{m/s} \]

4. **Use the Relationship Between Horizontal and Vertical Components:**
- The vertical component of the velocity when the ball makes an angle \( 45^\circ \) is given by:
\[ v_y = v \sin \phi \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ v_y = v \cdot \frac{1}{\sqrt{2}} \]

5. **Use the Pythagorean Theorem to Find the Magnitude of Velocity \( v \):**
- The total velocity \( v \) can be expressed in terms of its components:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting \( v_y \):
\[ v = \sqrt{(15)^2 + \left(v \cdot \frac{1}{\sqrt{2}}\right)^2} \]
Simplifying:
\[ v = \sqrt{225 + \frac{v^2}{2}} \]

6. **Square Both Sides to Eliminate the Square Root:**
\[ v^2 = 225 + \frac{v^2}{2} \]
Multiply through by 2 to eliminate the fraction:
\[ 2v^2 = 450 + v^2 \]
Rearranging gives:
\[ v^2 - 450 = 0 \]
Thus:
\[ v^2 = 450 \]
Taking the square root:
\[ v = \sqrt{450} = 15\sqrt{2} \, \text{m/s} \]

### Final Answer:
The speed of the ball when it makes an angle of \( 45^\circ \) with the horizontal is \( 15\sqrt{2} \, \text{m/s} \).

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