Question

A particle is projected from horizontal making an angle of 53⁰ with initial velocity of 100 $m s^{- 1}$ . The time taken by the particle to make angle 45° from horizontal is

Moderate

Solution

Component (100 cos 53⁰) 60 $m s^{- 1}$ will remain unchanged.
Velocity will make 45⁰ with horizontal when vertical component also becomes ± 60 $m s^{- 1}$ .
Using $v = u + a t$ (In vertical direction)
$+ 60 = 80 + (- 10) t_{1}$
$∴ t_{1} = 2 s \Rightarrow - 60 = 80 + (- 10) t_{2}$
$∴ t_{2} = 14 s$

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