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

Options

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

Answer

To solve the problem of a body dropped from a balloon that is rising with a uniform velocity \( u \), we will use the equations of motion. Here’s the step-by-step solution:

### Step 1: Understand the Initial Conditions
- The balloon rises with a uniform velocity \( u \).
- When the body is dropped from the balloon, it initially has an upward velocity \( u \) (the same as the balloon's velocity).
- The body will experience downward acceleration due to gravity, denoted as \( g \).

### Step 2: Define the Variables
- Let the height from which the body is dropped be \( h \).
- The initial velocity \( v_0 = u \) (upward).
- The acceleration \( a = -g \) (downward).
- The displacement \( s = -h \) (since the body moves downward).

### Step 3: Apply the Equation of Motion
We will use the second equation of motion:
\[ s = v_0 t + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ -h = u t - \frac{1}{2} g t^2 \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{1}{2} g t^2 - u t - h = 0 \]
This is a quadratic equation in the form of:
\[ at^2 + bt + c = 0 \]
where:
- \( a = \frac{1}{2} g \)
- \( b = -u \)
- \( c = -h \)

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-(-u) \pm \sqrt{(-u)^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g} \]
\[ t = \frac{u \pm \sqrt{u^2 + 2gh}}{g} \]

### Step 6: Determine the Time of Descent
Since time cannot be negative, we take the positive root:
\[ t = \frac{u + \sqrt{u^2 + 2gh}}{g} \]

### Final Answer
The time of descent for the body is given by:
\[ t = \frac{u + \sqrt{u^2 + 2gh}}{g} \]

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