A ball A is projected from the ground such that its horizontal range is maximum. Another ball B is dropped from a height equal to the maximum range of A. The ratio of the time of flight of A to the time of fall of B is

Options

A. `sqrt2:1`
B. `1:2`
C. `1:1`
D. `1:sqrt2`

Answer

To solve the problem step by step, we will analyze the motion of both balls A and B.

### Step 1: Determine the conditions for maximum range of ball A
To achieve maximum horizontal range, ball A must be projected at an angle of 45 degrees.

### Step 2: Calculate the horizontal range of ball A
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For \( \theta = 45^\circ \):
\[ \sin(90^\circ) = 1 \]
Thus, the range becomes:
\[ R = \frac{u^2}{g} \]

### Step 3: Identify the height from which ball B is dropped
Ball B is dropped from a height equal to the maximum range of ball A, which we found to be:
\[ h = R = \frac{u^2}{g} \]

### Step 4: Calculate the time of flight of ball A
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]
For \( \theta = 45^\circ \):
\[ T = \frac{2u \cdot \frac{\sqrt{2}}{2}}{g} = \frac{u \sqrt{2}}{g} \]

### Step 5: Calculate the time of fall of ball B
For ball B, which is dropped from height \( h \), we can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \), \( s = h = \frac{u^2}{g} \), and \( a = g \):
\[ \frac{u^2}{g} = 0 + \frac{1}{2} g t^2 \]
This simplifies to:
\[ \frac{u^2}{g} = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2u^2}{g^2} \]
Taking the square root:
\[ t = \frac{u \sqrt{2}}{g} \]

### Step 6: Find the ratio of the time of flight of A to the time of fall of B
Let \( T_A \) be the time of flight of ball A and \( T_B \) be the time of fall of ball B:
\[ T_A = \frac{u \sqrt{2}}{g}, \quad T_B = \frac{u \sqrt{2}}{g} \]
Thus, the ratio \( \frac{T_A}{T_B} \) is:
\[ \frac{T_A}{T_B} = \frac{\frac{u \sqrt{2}}{g}}{\frac{u \sqrt{2}}{g}} = 1 \]

### Final Answer
The ratio of the time of flight of A to the time of fall of B is:
\[ \text{Ratio} = 1:1 \]