Question

A projectile is thrown in the upward direction making an angle of 60${}^{\circ }$ with the horizontal direction with a velocity of 147 $m{s}^{-1}$ 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)

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 = $\frac{147}{\sqrt{2}}m{s}^{-1}$
Vertical component of u = usin 60${}^{\circ }$
$\frac{147\sqrt{3}}{2}m$
Vertical component of v = vsin 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 9.8 t = $\frac{147}{2}(\sqrt{3}-1)$
$\therefore t=5.49$ s