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 (Take $g=9.8m/s^2$)

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

Answer 1

The correct option is C $5.49s$

velocity of projectile$=u=147m{s}^{-1}$ angle of projection $\alpha ={60}^{\circ }$
Let the time taken by projectile from $O$ to $A$, be $t$ and its velocity is $v$ at point A where direction is $\beta ={45}^{\circ }$. As horizontal component of velocity remains constant during the projectile projectile motion.

$\Rightarrow v\cos {45}^{\circ }=u\cos {60}^{\circ }$

$\Rightarrow v\times \frac{1}{\sqrt{2}}=147\times \frac{1}{2}\Rightarrow v=\frac{147}{\sqrt{2}}m{s}^{-1}$

For vertical motion,
$v_y=u_y-gt$
$\Rightarrow v\sin {45}^{\circ }=45\sin {60}^{\circ }-9.8t$

$\Rightarrow \frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=147\times \frac{\sqrt{3}}{2}-9.8t$

$9.8t=\frac{147}{2}(\sqrt{3}-1)$
$t=5.49s$