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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Motion in One Dimension

AIIMS 2018 · 27 May · Physics · Motion in One Dimension

A particle is thrown vertically up with an initial velocity 9 ~m /...

A particle is thrown vertically up with an initial velocity 9 m/s9 \mathrm{~m} / \mathrm{s} from the surface of Earth (take g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 ). The time (in s) taken by the particle to reach a height of 4 m4 \mathrm{~m} from the surface second time (in seconds) is

Options

  1. A1.31.3
  2. B1.21.2
  3. C1.11.1
  4. D1.01.0

Correct answer

D. 1.01.0

Step-by-step solution

Time taken by the particle to reach a height hh from the surface of Earth is obtained from h=ut−12gt2h=u t-\frac{1}{2} g t^2 Here, u=9 m/s,h=4 m,g=10 m/s2u=9 \mathrm{~m} / \mathrm{s}, h=4 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2∴4=9t−12×10×t24=9t−5t2 or 5t2−9t+4=05t2−5t−4t+4=0⇒5t(t−1)−4(t−1)=0∴t=1 s or t=45 s\begin{aligned} & \therefore \quad 4=9 t-\frac{1}{2} \times 10 \times t^2 \\ & 4=9 t-5 t^2 \quad \text{ or } 5 t^2-9 t+4=0 \\ & 5 t^2-5 t-4 t+4=0 \\ & \Rightarrow \quad 5 t(t-1)-4(t-1)=0 \\ & \therefore \quad t=1 \mathrm{~s} \text{ or } t=\frac{4}{5} \mathrm{~s}\end{aligned}$Hence,thetimetakenbytheparticletoreachaheightofHence, the time taken by the particle to reach a height of4 \mathrm{~m}fromthesurfacesecondtimeisfrom the surface second time is1 \mathrm{~s}$.
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