MCQ
Particle is dropped from the height of $20\,\,m$ from horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes $6 ms^{^{-2}}$. Find the horizontal displacement of the particle till it reaches ground.  ........ $m$
  • A
    $6 $
  • B
    $10$
  • $12$
  • D
    $24$

Answer

Correct option: C.
$12$
c
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 20}{10}}=2 \mathrm{s}$

$v_{x}=0$

$a _x=6 m / s^{2}$

$t=25$

$R=v_{x} t+\frac{1}{2} a_{x} t^{2}$

$=0+\frac{1}{2} \times 6 \times 2^{2}$

$R=12 \mathrm{m}$

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