Question

A parachutist after bailing out falls $50m$ without friction. When the parachute opens, it decelerates at $2m{{s}^{-2}}$. He reaches the ground with speed $3m{{s}^{-2}}$. At what height did he bail out?

Answer 1

Hint: In this use basic equations of motion that describes the nature of any kind of physical system in terms of its motion which in turn is a function of time. As we can see when a parachutist jumps so when he is in the air the acceleration will be constant due to gravity. So, after the jump first calculate the initial velocity and then calculate the height with the help of initial velocity and final velocity after reaching the ground.

Formula used:
$u=\sqrt{2gh}$
${{v}^{2}}-{{u}^{2}}=2as$
Where: $v$= final velocity
$u$= initial velocity
$s$= distance

Complete step by step answer:
Now, as given in problem that the height = $50m$
And as the parachutist jump so calculate initial velocity with the help of height from which he jumped
So,
Initial velocity, $u=\sqrt{2gh}$
Or, $u=\sqrt{2\times 9.8\times 50}=14\sqrt{5}m/\operatorname{s}$
The velocity at ground, $v=3{m}/{s}\;$
$2as={{v}^{2}}-{{u}^{2}}$
Or, $2\times s\times \left( -2 \right)={{3}^{2}}-980$
So,
Height $s=\dfrac{971}{4}\approx 243m$
Total height at the parachutist bailed out = $243m+50m=293m$
Hence, the correct answer is option C.

Additional Information
The basic equations of motion describes the fundamental concepts of motion of any object. These equations direct the motion of an object in 1, 2, and 3 dimensions. These equations are used to derive the components like velocities, displacement, and acceleration of any object. therefore, they can only be used when motion is a straight line and acceleration is constant.

Note:
The signs on the quantities should be the same if they point in the same direction and opposite if they point in the opposite direction. So basically, choose as per your concern. Like if you are using the acceleration due to gravity positive in downward direction then choose the signs for other data’s as per sign convention of acceleration due to gravity that you are using.