Question

An open lift is coming down from the top of a building at a constant speed $\mathrm{v}=10$ $\mathrm{m} / \mathrm{s} .$ A boy standing on the lift throws a stone vertically upwards at a speed 30 $\mathrm{m} / \mathrm{s} .$ w.r.t. himself. The time after which he will catch the stone is
A.4 sec
B.6 sec
C.8 sec
D.10 sec

Answer 1

Hint:Formula used: $\text{V=u+at}$

Complete answer:
Given in the question
Speed of lift vertically downwards $=10 \mathrm{~m} / \mathrm{~s}$
Stone speed w.r.t boy vertically upwards $=30 \mathrm{~m} / \mathrm{~s}$
Stone speed w.r.t lift vertically upwards
$\Rightarrow 30-10$
$=20~\text{m}/~\text{s}$
As boy is standing in the lift his velocity w.r.t ground $=10 \mathrm{~m} / \mathrm{~s}$
Stone velocity $w$ r.t ground
$\Rightarrow 20+10$
$=30~\text{m}/~\text{s}$
As stone reaches maximum height velocity of stone $=0$
From the equation of the motion
$V=U-a t$
$\Rightarrow 0=30-10(t)$
$\Rightarrow \text{t}=3\text{sec}$ to reach maximum height
$\Rightarrow 3+3$ seconds
$=6\text{s}$
Total time Taken to catch the stone is

The correct option is (b).

Note:
Under these key classifiers of motion, equations of motion may also be categorized. In all cases, translations, rotations, oscillations, or some combination of these are the main forms of motion. For a particle with constant or uniform acceleration in a straight line, the differential equation of motion is simple: the acceleration is constant, so the second derivative of the object's location is constant. Typically, these variables are spatial coordinates and time, but can include components of momentum. Generalized coordinates are the most common choice, which can be any convenient variables that describe the physical system.