A stone is thrown at `25m//s " at " 53^(@)` above the horizontal. At what time its velocity is at `45^(@)` below the horizontal?

Options

A. 0.5s
B. 4s
C. 3.5s
D. 2.5s

Answer

To solve the problem, we need to analyze the motion of the stone thrown at an angle. We'll break it down step by step.

### Step 1: Determine the initial velocity components
The stone is thrown with an initial velocity \( u = 25 \, \text{m/s} \) at an angle \( \theta = 53^\circ \) above the horizontal.

- The horizontal component of the initial velocity (\( u_x \)):
\[ u_x = u \cos(\theta) = 25 \cos(53^\circ) \]
Using \( \cos(53^\circ) \approx \frac{3}{5} \):
\[ u_x = 25 \times \frac{3}{5} = 15 \, \text{m/s} \]

- The vertical component of the initial velocity (\( u_y \)):
\[ u_y = u \sin(\theta) = 25 \sin(53^\circ) \]
Using \( \sin(53^\circ) \approx \frac{4}{5} \):
\[ u_y = 25 \times \frac{4}{5} = 20 \, \text{m/s} \]

### Step 2: Analyze the velocity at 45 degrees below the horizontal
At the time we want to find, the stone's velocity is at \( 45^\circ \) below the horizontal. This means the horizontal and vertical components of the velocity are equal in magnitude but opposite in direction.

Let \( V \) be the magnitude of the velocity at this point. The horizontal component (\( V_x \)) is:
\[ V_x = V \cos(45^\circ) = V \frac{1}{\sqrt{2}} \]
And since the horizontal component remains constant:
\[ V_x = u_x = 15 \, \text{m/s} \]

From this, we can find \( V \):
\[ V \frac{1}{\sqrt{2}} = 15 \implies V = 15 \sqrt{2} \, \text{m/s} \]

### Step 3: Find the vertical component of the velocity
The vertical component of the velocity when the stone is at \( 45^\circ \) below the horizontal is:
\[ V_y = -V \sin(45^\circ) = -V \frac{1}{\sqrt{2}} \]
Substituting \( V \):
\[ V_y = -15 \sqrt{2} \frac{1}{\sqrt{2}} = -15 \, \text{m/s} \]

### Step 4: Use the kinematic equation to find the time
We can use the kinematic equation for vertical motion:
\[ V_y = u_y - g t \]
Substituting the known values:
\[ -15 = 20 - 10t \]
Rearranging gives:
\[ 10t = 20 + 15 \implies 10t = 35 \implies t = 3.5 \, \text{s} \]

### Final Answer
The time at which the stone's velocity is at \( 45^\circ \) below the horizontal is \( t = 3.5 \, \text{s} \).

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