Question

A projectile is thrown in the upward direction making an angle of $60^{\circ }$ with the horizontal direction with a velocity of $150m{s}^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ }$ is

• A. $15(\sqrt{3}-1)s$
• B. $15(\sqrt{3}+1)s$
• C. $7.5(\sqrt{3}-1)s$
• D. $7.5(\sqrt{3}+1)s$

Answer 1

Answer: (C)

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then
$150\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}$
or $v=\frac{150}{\sqrt{2}}m{s}^{-1}$
Initially: $u_y=u\sin 60^{\circ }=\frac{150\sqrt{3}}{2}m{s}^{-1}$
Finally: $v_y=v\sin 45^{\circ }=\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}m{s}^{-1}$
But $v_y=u_y+a_{y}t$
or $\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t$
$10t=\frac{150}{2}(\sqrt{3}-1)$
or $t=7.5(\sqrt{3}-1)$