Question

A parachutist jumps from a helicopter. It falls freely for 2 sec. Then he opens parachute which produces retardation of 3 m/s\(^2\). When his height from ground is 10 m his velocity is 5 m/s. Find his initial height from ground.

• A. 90 m
• B. 82 m
• C. 92.5 m
• D. 100 m

Hint:

For multi-stage motion problems, it is crucial to clearly define the initial and final conditions for each stage. The final velocity of one stage becomes the initial velocity for the next. Draw a simple diagram to visualize the different parts of the journey.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The motion of the parachutist is divided into two distinct phases of constant acceleration, followed by a final position. We need to find the total initial height by summing the distances covered in each phase and the final height above the ground. We will assume \(g = 10 \text{ m/s}^2\).
Step 2: Key Formula or Approach:
We will use the equations of motion for constant acceleration:
- \(v = u + at\)
- \(s = ut + \frac{1}{2}at^2\)
- \(v^2 = u^2 + 2as\)
Step 3: Detailed Explanation:
Phase 1: Free Fall
- Initial velocity, \(u_1 = 0\).
- Time, \(t_1 = 2\) s.
- Acceleration, \(a_1 = g = 10 \text{ m/s}^2\).
The distance fallen during this phase is \(S_1\):
\[ S_1 = u_1t_1 + \frac{1}{2}a_1t_1^2 = 0 + \frac{1}{2}(10)(2)^2 = 20 \text{ m} \] The velocity at the end of this phase is \(v_1\):
\[ v_1 = u_1 + a_1t_1 = 0 + (10)(2) = 20 \text{ m/s} \] Phase 2: Parachute Open (Retardation)
- Initial velocity for this phase, \(u_2 = v_1 = 20 \text{ m/s}\).
- Acceleration, \(a_2 = -3 \text{ m/s}^2\) (retardation).
- Final velocity for this phase, \(v_2 = 5 \text{ m/s}\).
The distance fallen during this phase is \(S_2\). We use the third equation of motion:
\[ v_2^2 = u_2^2 + 2a_2S_2 \] \[ (5)^2 = (20)^2 + 2(-3)S_2 \] \[ 25 = 400 - 6S_2 \] \[ 6S_2 = 400 - 25 = 375 \] \[ S_2 = \frac{375}{6} = 62.5 \text{ m} \] Total Initial Height
This second phase ends when the parachutist is at a height of 10 m from the ground. The total initial height (\(H\)) is the sum of the distances fallen and this final height.
\[ H = S_1 + S_2 + \text{Final Height} \] \[ H = 20 \text{ m} + 62.5 \text{ m} + 10 \text{ m} = 92.5 \text{ m} \] Step 4: Final Answer:
The initial height of the parachutist from the ground was 92.5 m.