A parachutist drops first freely form an areophone for `10 s` and then his parachut opens out. Now he descends with a net retardtion of `2.5 ms^(-2)` If the balil out of the plane at a height of `2495 m` and `g=10 ms^(-2)`, his velocity on reaching the ground will be`.

Options

A. `5 m s^(-1)`
B. `10 ms^(-1)`
C. `15 ms^-1)`
D. `20 ms^(-1)`

Explanation

To solve the problem step by step, we will break it down into parts:

### Step 1: Calculate the distance fallen during the first 10 seconds
The parachutist falls freely for 10 seconds. The distance fallen under gravity can be calculated using the second equation of motion:

\[ s = ut + \frac{1}{2} g t^2 \]

Here,
- \( u = 0 \) (initial velocity, as the parachutist starts from rest),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t = 10 \, \text{s} \).

Substituting the values:

\[ s = 0 \cdot 10 + \frac{1}{2} \cdot 10 \cdot (10)^2 = \frac{1}{2} \cdot 10 \cdot 100 = 500 \, \text{m} \]

### Step 2: Calculate the velocity just before the parachute opens
The velocity at the end of 10 seconds can be calculated using the first equation of motion:

\[ v = u + gt \]

Substituting the values:

\[ v = 0 + 10 \cdot 10 = 100 \, \text{m/s} \]

### Step 3: Calculate the height from which the parachutist descends after the parachute opens
The parachutist starts at a height of 2495 m. After falling 500 m, the height remaining when the parachute opens is:

\[ \text{Remaining height} = 2495 \, \text{m} - 500 \, \text{m} = 1995 \, \text{m} \]

### Step 4: Calculate the final velocity when reaching the ground
Now, the parachutist descends with a net retardation of \( 2.5 \, \text{m/s}^2 \). We can use the third equation of motion to find the final velocity when he reaches the ground:

\[ v^2 = u^2 + 2as \]

Here,
- \( u = 100 \, \text{m/s} \) (initial velocity when the parachute opens),
- \( a = -2.5 \, \text{m/s}^2 \) (retardation, hence negative),
- \( s = 1995 \, \text{m} \) (distance to the ground).

Substituting the values:

\[ v^2 = (100)^2 + 2 \cdot (-2.5) \cdot 1995 \]

Calculating:

\[ v^2 = 10000 - 2 \cdot 2.5 \cdot 1995 \]
\[ v^2 = 10000 - 9975 = 25 \]
\[ v = \sqrt{25} = 5 \, \text{m/s} \]

### Final Answer
The velocity of the parachutist when reaching the ground is \( 5 \, \text{m/s} \).

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