Animated Solution for Physics - Kinematics: A parachutist after bailing out falls $50\text{ m}$ without friction. When parachute opens, it decelerates at $2{\text{ ms}}^{-2}$. He reaches the ground with a speed of $3{\text{ ms}}^{-1}$. At what height, did he bail out?

Imagine you are standing at the edge of an airplane door, looking down at the earth far below. You take a deep breath and jump! This thrilling scenario is exactly what we are analyzing in this problem. To solve it, we need to break the parachutist's journey into two distinct, manageable phases.

The Free Fall

For the first $50\text{ m}$, you are in a state of free fall. The problem states there is no air friction during this phase, which means the only force acting on you is gravity. You start from rest, so your initial velocity $u=0$. The acceleration is due to gravity, $a=g=9.8{\text{ m/s}}^2$, and the distance covered is $h=50\text{ m}$.

We need to find out how fast you are going right before you pull the parachute cord. The perfect tool for this is the third equation of motion, which relates velocity, acceleration, and distance without needing time:

Substituting our known values:

Pro Tip: Notice how we didn't calculate the square root of $980$? That's a strategic move! We know we will need $v^2$ in the next phase, so leaving it as $980$ saves us from messy decimal calculations.

The Decelerated Descent

Now, you pull the cord! The parachute blossoms open, and you experience a sudden upward jerk. You are now decelerating. The final velocity of Phase 1 becomes the initial velocity for Phase 2. So, our new initial velocity squared is $v^2=980$.

During this phase, the parachute slows you down with an acceleration of $a=-2{\text{ m/s}}^2$. The negative sign is crucial here because the acceleration is acting opposite to your downward motion. You finally reach the ground with a safe landing speed of $v_1=3\text{ m/s}$. Let's call the distance covered in this phase $h^{\prime }$.

We apply the third equation of motion once more:

Substituting the values for this phase:

Now, it's just a matter of simple algebra to isolate $h^{\prime }$:

We have successfully calculated the distances for both phases of the jump. To find the total height $H$ from which you bailed out, we simply add the free-fall distance to the decelerated distance:

Looking at our options, $292.75\text{ m}$ is incredibly close to $293\text{ m}$. Therefore, the correct option is (c). What an exhilarating application of kinematics!