A object is thrown horizontally from a point A. The ball hits the ground at a point B. The line of sight from A to B makes 60° with the horizontal. If g = `10ms^(-2)`, the velocity of projection, if the time of flight is `sqrt(3)` sec. is

Options

A. `10 ms^(-1)`
B. `5 ms^(-1)`
C. `10sqrt(3)ms^(-1)`
D. `10/sqrt(3)ms^(-1)`

Answer

To solve the problem step by step, we will analyze the motion of the object thrown horizontally from point A and hitting the ground at point B.

### Step 1: Understanding the Problem
The object is thrown horizontally from point A and hits the ground at point B, making a line of sight at an angle of 60° with the horizontal. We are given the time of flight \( t = \sqrt{3} \) seconds and the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 2: Determine the Vertical Distance (y)
Using the equation of motion for vertical displacement:
\[ y = ut + \frac{1}{2} a t^2 \]
Since the object is thrown horizontally, the initial vertical velocity \( u = 0 \). Thus, the equation simplifies to:
\[ y = 0 + \frac{1}{2} (-g) t^2 \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( t = \sqrt{3} \):
\[ y = \frac{1}{2} (-10) (\sqrt{3})^2 = \frac{1}{2} (-10) \cdot 3 = -15 \, \text{m} \]
The negative sign indicates that the displacement is downward, so the vertical distance \( y = 15 \, \text{m} \).

### Step 3: Relate y and x using the Angle
From the geometry of the problem, we know that:
\[ \tan(60^\circ) = \frac{y}{x} \]
Substituting the known value of \( y \):
\[ \tan(60^\circ) = \sqrt{3} \quad \Rightarrow \quad \sqrt{3} = \frac{15}{x} \]
Rearranging gives:
\[ x = \frac{15}{\sqrt{3}} = 5\sqrt{3} \, \text{m} \]

### Step 4: Calculate the Horizontal Velocity (v)
The horizontal distance \( x \) is covered in the time of flight \( t \):
\[ x = v \cdot t \]
Substituting \( x = 5\sqrt{3} \) and \( t = \sqrt{3} \):
\[ 5\sqrt{3} = v \cdot \sqrt{3} \]
Dividing both sides by \( \sqrt{3} \):
\[ v = 5 \, \text{m/s} \]

### Final Answer
The velocity of projection is \( \boxed{5 \, \text{m/s}} \).

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