MCQ
A particle is projected from horizontal making an angle of $53^{\circ}$ with initial velocity $100\,m / s$. The time taken by the particle to make angle $45^{\circ}$ from horizontal is $.........\,s$
  • A
    $14$
  • B
    $2$
  • $(a)$ and $(b)$ both
  • D
    $4$

Answer

Correct option: C.
$(a)$ and $(b)$ both
c
(c)

Component $60\,m / s$ will remain unchanged. Velocity will make $45^{\circ}$ with horizontal when vertical component also becomes $\pm 60\,m / s$.

Using,$v=u+a t$ (in vertical direction)

$+60=80+(-10) t_1$

$\therefore \quad t_1=2\,s$

$-60=80+(-10) t_2$

$\therefore \quad t_2=14\,s$

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