A stone projected at angle `53^(@)` attains maximum height 25m during its motion in air. Then its distance from the point of projection where it will fall is

Options

A. 75m
B. `400/3 m`
C. `50 m`
D. `60 m`

Explanation

To solve the problem step by step, we will use the principles of projectile motion. The stone is projected at an angle of \(53^\circ\) and reaches a maximum height of \(25\) meters. We need to find the horizontal distance (range) from the point of projection to where the stone will land.

### Step 1: Understanding the Components of Motion
When the stone is projected, it has two components of motion: horizontal and vertical. The initial velocity \(V\) can be broken down into:
- Horizontal component: \(V_x = V \cos(53^\circ)\)
- Vertical component: \(V_y = V \sin(53^\circ)\)

### Step 2: Finding the Initial Vertical Velocity
At the maximum height, the final vertical velocity is \(0\). We can use the kinematic equation:
\[ V_y^2 = U_y^2 + 2a s \]
Where:
- \(V_y = 0\) (final vertical velocity at maximum height)
- \(U_y = V \sin(53^\circ)\) (initial vertical velocity)
- \(a = -g\) (acceleration due to gravity, negative because it acts downwards)
- \(s = 25\) m (maximum height)

Substituting the values:
\[ 0 = (V \sin(53^\circ))^2 - 2g(25) \]
\[ (V \sin(53^\circ))^2 = 50g \]

### Step 3: Calculating \(\sin(53^\circ)\)
Using the value of \(\sin(53^\circ) = \frac{4}{5}\):
\[ (V \cdot \frac{4}{5})^2 = 50g \]
\[ \frac{16V^2}{25} = 50g \]
\[ V^2 = \frac{50g \cdot 25}{16} \]

### Step 4: Finding the Total Time of Flight
The time to reach the maximum height can be calculated using:
\[ t = \frac{V_y}{g} = \frac{V \sin(53^\circ)}{g} \]
Substituting \(\sin(53^\circ)\):
\[ t = \frac{V \cdot \frac{4}{5}}{g} = \frac{4V}{5g} \]
The total time of flight \(T\) is twice this time:
\[ T = 2t = \frac{8V}{5g} \]

### Step 5: Finding the Range
The range \(R\) can be calculated using the horizontal velocity and the total time of flight:
\[ R = V_x \cdot T = (V \cos(53^\circ)) \cdot T \]
Substituting \(\cos(53^\circ) = \frac{3}{5}\):
\[ R = V \cdot \frac{3}{5} \cdot \frac{8V}{5g} \]
\[ R = \frac{24V^2}{25g} \]

### Step 6: Substituting \(V^2\) into the Range Formula
Now substitute \(V^2\) from earlier:
\[ R = \frac{24 \cdot \frac{50g \cdot 25}{16}}{25g} \]
\[ R = \frac{24 \cdot 50 \cdot 25}{16 \cdot 25} \]
\[ R = \frac{1200}{16} = 75 \text{ meters} \]

### Final Answer
The distance from the point of projection where the stone will fall is \(75\) meters.

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