Q125 / 180 · AIIMS 2018 (27 May) · Physics

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

  • A
    $1.3$
  • B
    $1.2$
  • C
    $1.1$
  • Correct answer
    D
    $1.0$

Solution

Time taken by the particle to reach a height $h$ from the surface of Earth is obtained from $h=u t-\frac{1}{2} g t^2$ Here, $u=9 \mathrm{~m} / \mathrm{s}, h=4 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$ $\begin{aligned} & \therefore \quad 4=9 t-\frac{1}{2} \times 10 \times t^2 \\ & 4=9 t-5 t^2 \quad \text { or } 5 t^2-9 t+4=0 \\ & 5 t^2-5 t-4 t+4=0 \\ & \Rightarrow \quad 5 t(t-1)-4(t-1)=0 \\ & \therefore \quad t=1 \mathrm{~s} \text { or } t=\frac{4}{5} \mathrm{~s}\end{aligned}$ Hence, the time taken by the particle to reach a height of $4 \mathrm{~m}$ from the surface second time is $1 \mathrm{~s}$.

Hence, the correct answer is option: 'D'.

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