Question

A parachutist, after bailing out, falls 50 m without friction. When parachute opens, it retards at 2 ms^-2. He reaches the ground with a speed of 3 ms^-1. At what height did he bail out?

• (A) 293 m
• (B) 111 m
• (C) 91 m
• (D) 182 m

Answer 1

The Correct Option is A

Explanation:
Let velocity at the end of $50\mathrm{m}$ be u $\mathrm{m}/\mathrm{s}$$$u^2=2gh=2\times 9.8\times 50=980\text{ units }$$Let further displacement be s.$$s=\frac{v^2-u^2}{2a}-\frac{3^2-980}{-2\times 2}$$$s=\frac{980-9}{4}=\frac{971}{4}=243$$\therefore$ Total height $=h+s=50+243=293\mathrm{m}$

Answer 2

A mathematical formula that defines how a physical system acts over time is known as an equation of motion. The equation of motion defines the motion of objects and systems in terms of dynamic variables. These equations connect many important motion characteristics such as velocity, displacement, speed, time, and acceleration.

Equations of Motion

There are three equations of motion that are only applicable to uniformly accelerated motion. The following are the equations:

1. v = u + at
2. v² = u² + 2as
3. s = ut + 1/2 at²

Where

• v is the final Velocity 
• u is the initial velocity 
• t is the time taken 
• a is constant acceleration or uniform acceleration
• s is the displacement of the body

Equations of Motion Under Gravity

When the motion of an object is due to the action of gravity, then equations of motion are followed for the analysis of the motion of the object. These equations are

1. v = u + gt
2. v² = u² + 2gs
3. s = ut + 1/2 gt²

Where g is acceleration due to gravity. There is a sign convention followed for the acceleration due to gravity in the above equations. It is taken as positive for a downward direction and negative for an upward direction.