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)
(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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