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Step-by-Step Solution
Step 1: Identify the two phases of motion
1. Free fall without friction for a distance of $50\,\text{m}$.
2. Decelerated motion (when the parachute opens) until the parachutist reaches the ground with a final speed of $3\,\text{m/s}$.
Step 2: Calculate the velocity after the first (free-fall) phase
Since the parachutist starts from rest and falls $50\,\text{m}$ under gravity, we use the kinematic equation:
$ v^2 = u^2 + 2 g h $
where:
$u = 0\,\text{m/s}$ (initial velocity)
$g = 9.8\,\text{m/s}^2$ (acceleration due to gravity)
$h = 50\,\text{m}$
Thus,
$ v^2 = 0^2 + 2 \times 9.8 \times 50 $
$ v^2 = 980 $
$ v = \sqrt{980} \approx 31.3\,\text{m/s} $
So, at the end of this phase, the parachutist’s velocity is approximately $31.3\,\text{m/s}$ downward.
Step 3: Set up the decelerated motion (second phase)
After the parachute opens, the parachutist slows down with a uniform deceleration of $2\,\text{m/s}^2$. He continues until his speed reduces from $31.3\,\text{m/s}$ to $3\,\text{m/s}$.
We use the kinematic equation:
$ v^2 = u^2 + 2\,a\,s $
Here, since the motion is decelerated, the acceleration $a$ is $-2\,\text{m/s}^2$. Let $s$ be the distance covered in this phase:
$ (3)^2 = (31.3)^2 + 2 \times (-2)\times s $
$ 9 = 31.3^2 - 4s $
$ 4s = 31.3^2 - 9 $
$ 4s = 979.69 - 9 $
$ 4s = 970.69 $
$ s = \frac{970.69}{4} \approx 242.67\,\text{m} $
Step 4: Sum up the total height
The total height from which the parachutist bailed out is the sum of the distance covered during free fall and the distance covered during the decelerated motion:
$ H_\text{total} = 50\,\text{m} + 242.67\,\text{m} \approx 293\,\text{m}. $
Final Answer
The parachutist bailed out from a height of approximately $293\,\text{m}$.
