A body falling from rest under gravity passes a certain point P . It was at a distance of 400 ~m from P, 4 ~s prior to passing through P . If g=10 ~m / s^2 , then the height above the point P from where the body began

Correct answer: A. 720 m We have h=(1)/(2) g t^2 ^2 and h+400=(1)/(2) g(t+4)^2 . Subtracting we get 400=8 g+4 gt aligned & t=8 sec & h=(1)/(2) × 10 × 64=320 m & Desired height =320+400=720 m . aligned

JEE Main 2006

Exam: JEE Main
Year: 2006
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2006 · Physics · Motion in One Dimension

A body falling from rest under gravity passes a certain point P ....

A body falling from rest under gravity passes a certain point

P

. It was at a distance of

400 \mathrm{~m}

from

P, 4 \mathrm{~s}

prior to passing through

P

. If

g=10 \mathrm{~m} / \mathrm{s}^2

, then the height above the point

\mathrm{P}

from where the body began to fall is

Options

A720 m
B900 m
C320 m
D680 m

Correct answer

A. 720 m

Step-by-step solution

We have

h=\frac{1}{2}{g t^2}^2

and

h+400=\frac{1}{2} g(t+4)^2

. Subtracting we get

400=8 \mathrm{ g}+4 \mathrm{gt}

\begin{aligned} & \Rightarrow \mathrm{t}=8 \mathrm{sec} \\ & \therefore \mathrm{h}=\frac{1}{2} \times 10 \times 64=320 \mathrm{ m} \\ & \therefore \text{ Desired height }=320+400=720 \mathrm{ m} . \end{aligned}

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