A ball is thrown with a speed u, at an angle `theta` with the horizontal. At the highest point of its motion, the strength of gravity is somehow doubled. Taking this change into account, the total time of flight of the projectile is

Options

A. `(2u sin theta)/g`
B. `3/2 (u sin theta)/g`
C. `3/4 (u sin theta)/g`
D. `((sqrt(2)+1)/sqrt(2)) (u sin theta)/g`

Answer

To solve the problem, we need to determine the total time of flight of a projectile when the strength of gravity is doubled at the highest point of its motion. Let's break this down step by step.

### Step 1: Understanding the Initial Conditions
The ball is thrown with an initial speed \( u \) at an angle \( \theta \) with the horizontal. The initial vertical and horizontal components of the velocity can be expressed as:
- \( u_y = u \sin \theta \) (vertical component)
- \( u_x = u \cos \theta \) (horizontal component)

### Step 2: Time to Reach the Highest Point
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero. The time taken to reach this highest point (let's denote it as \( T_1 \)) can be found using the formula:
\[ T_1 = \frac{u_y}{g} = \frac{u \sin \theta}{g} \]

### Step 3: Height at the Highest Point
The height \( h \) reached at the highest point can be calculated using the equation of motion:
\[ h = u_y T_1 - \frac{1}{2} g T_1^2 \]
Substituting \( T_1 \):
\[ h = u \sin \theta \left(\frac{u \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2 \]
\[ h = \frac{u^2 \sin^2 \theta}{g} - \frac{1}{2} g \cdot \frac{u^2 \sin^2 \theta}{g^2} \]
\[ h = \frac{u^2 \sin^2 \theta}{g} - \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 4: Change in Gravity at the Highest Point
At the highest point, the strength of gravity is doubled, so the new gravitational acceleration \( g' = 2g \).

### Step 5: Time to Fall from the Highest Point to the Ground
Now, we need to calculate the time taken to fall from the highest point back to the ground (let's denote this time as \( T_2 \)). Using the second equation of motion:
\[ h = \frac{1}{2} g' T_2^2 \]
Substituting \( g' \) and \( h \):
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} (2g) T_2^2 \]
\[ \frac{u^2 \sin^2 \theta}{2g} = g T_2^2 \]
\[ T_2^2 = \frac{u^2 \sin^2 \theta}{2g^2} \]
\[ T_2 = \frac{u \sin \theta}{\sqrt{2} g} \]

### Step 6: Total Time of Flight
The total time of flight \( T \) is the sum of the time taken to reach the highest point and the time taken to fall back to the ground:
\[ T = T_1 + T_2 \]
Substituting the values we found:
\[ T = \frac{u \sin \theta}{g} + \frac{u \sin \theta}{\sqrt{2} g} \]
Factoring out \( \frac{u \sin \theta}{g} \):
\[ T = \frac{u \sin \theta}{g} \left(1 + \frac{1}{\sqrt{2}}\right) \]

### Final Answer
Thus, the total time of flight of the projectile is:
\[ T = \frac{u \sin \theta}{g} \left(1 + \frac{1}{\sqrt{2}}\right) \]