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

Answer 1

Answer: (C)

Initial velocity of projectile, $u=147 \,m / s$

At the two points of the trajectory during projection, the horizontal component of the velocity is the same.
$u \cos 60^{\circ}=v \cos 45^{\circ}$
$147 \times \frac{1}{2}=v \times \frac{1}{\sqrt{2}} $
$v=\frac{147}{\sqrt{2}} m / s$
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 $
$v_{y}=u_{y}-g t $
$\frac{147}{2}=\frac{147 \sqrt{3}}{2}-9.8\, t$
$9.8 t=\frac{147}{2}(\sqrt{3}-1) $
$t=\frac{147}{2 \times 9.8}(\sqrt{3}-1)$
$=5.49\, s$