Question

A projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147\, ms ^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$, is

• A. 15 s
• B. 10.98 s
• C. 5.49 s
• D. 2.745 s

Answer 1

Answer: (C)

Key Idea At the two points of the trajectory during projection, the horizontal component of the velocity is the same.
Horizontal component of velocity at angle $60^{\circ}$
$=$ Horizontal component of velocity at $45^{\circ}$
ie, $u \cos 60^{\circ}=v \sin 45^{\circ}$
or $147 \times \frac{1}{2}=v \times \frac{1}{\sqrt{2}}$
or $v=\frac{147}{\sqrt{2}} m / s$
Vertical component of $u=u \sin 60^{\circ}$
$=\frac{147 \sqrt{3}}{2} m$
Vertical component of $v=v \sin 45^{\circ}$
$=\frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{147}{2} m$
but $ v_{y} =u_{y}+a t$
$ \therefore \frac{147}{2} =\frac{147 \sqrt{3}}{2}-9.8 t $
or $ 9.8 \,t =\frac{147}{2}(\sqrt{3}-1) $
$ \therefore t =5.49\, s $