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

Answer 1

Answer: (C)

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

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 $v=v\sin 45^{\circ }$
$=\frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}$
$=\frac{147}{2}m$
$v_y=u_y-gt$
$\frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8t$
$9.8t=\frac{147}{2}(\sqrt{3}-1)$
$t=\frac{147}{2\times 9.8}(\sqrt{3}-1)$
$=5.49s$