Question

A projectile is thrown in the upward direction making an angle of $60^{\circ }$ with the horizontal direction with a velocity of $147m{s}^{-1}$. 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 $v_y=u_y+at$
$\therefore \frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8t$
or $9.8t=\frac{147}{2}(\sqrt{3}-1)$
$\therefore t=5.49s$