A balloon rases up with uniform velocity 'u'. A body is dropped from ballon. The time of descent for the body is given by is

Options

A. `sqrt((2h)/g)`
B. `h=ut+1/2 g t^2`
C. `h=-ut +1/2 g t^2`
D. `-h=ut + 1/2 g t^2`

Answer

To solve the problem of a body being dropped from a balloon that rises with uniform velocity 'u', we need to analyze the motion of the body after it is released. Here’s a step-by-step solution:

### Step 1: Understand the scenario
The balloon is moving upwards with a uniform velocity 'u'. When the body is dropped from the balloon, it initially has the same upward velocity 'u' as the balloon.

### Step 2: Define the coordinate system
We will take the upward direction as positive. Therefore, the downward direction will be negative.

### Step 3: Identify the variables
- Initial velocity of the body (u) = +u (upward)
- Acceleration (a) = -g (downward)
- Displacement (s) = -h (the body moves downward to reach the ground)

### Step 4: Use the kinematic equation
We will use the kinematic equation for uniformly accelerated motion:
\[ s = ut + \frac{1}{2} a t^2 \]

### Step 5: Substitute the known values
Substituting the values we have:
- \( s = -h \)
- \( u = +u \)
- \( a = -g \)

The equation becomes:
\[ -h = ut + \frac{1}{2} (-g) t^2 \]

### Step 6: Simplify the equation
This simplifies to:
\[ -h = ut - \frac{1}{2} gt^2 \]

Rearranging gives:
\[ h = -ut + \frac{1}{2} gt^2 \]

### Step 7: Final form of the equation
Thus, we can express the equation as:
\[ h = -ut + \frac{1}{2} gt^2 \]

This indicates that the time of descent for the body can be determined from this equation, confirming that the correct option is:
\[ h = -ut + \frac{1}{2} gt^2 \]

### Conclusion
The time of descent for the body is given by the equation:
\[ h = -ut + \frac{1}{2} gt^2 \]