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A parachutist after bailing out falls 50m without friction. When the parachute opens, it decelerates at 2ms2. He reaches the ground with speed 3ms2. At what height did he bail out?

Answer
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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=2gh
v2u2=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=2gh
Or, u=2×9.8×50=145m/s
The velocity at ground, v=3m/s
2as=v2u2
Or, 2×s×(2)=32980
So,
Height s=9714243m
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.
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