Question

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

The correct option is C $7.5(\sqrt{3}-1)s$

At the two points of the trajectory during projectile motion, the horizontal component of velocity is same. Then,

$150\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\Rightarrow v=\frac{150}{\sqrt{2}}m{s}^{-1}$

Initially, $u_y=u\sin 60=\frac{150\sqrt{3}}{2}m{s}^{-1}$

Finally, $v_y=v\sin {45}^o=\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}m{s}^{-1}$

But $v_y=u_y+a_{y}t$

$\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t$

$\Rightarrow 10t=\frac{150(\sqrt{3}-1)}{2}$

$\Rightarrow t=7.5(\sqrt{3}-1)s$