# solve Q 9

Let’s assume that initial velocity of the particle is

**u**= u cosθ i + u sin θ j

Here, positive j axis is taken as vertically upward.

In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant.

Therefore let at point where particle will move perpendicular to its initial direction velocity is v

So,

**v**= u cosθ i + (u sin θ -gt) j (As vertical component changes according to kinematics equation v = u -gt)

Now as at this point as particle is moving perpendicular to initial direction so

$\overrightarrow{u}.\overrightarrow{v}=0\phantom{\rule{0ex}{0ex}}\left(u\mathrm{cos}\theta \hat{i}+u\mathrm{sin}\theta \hat{j}\right).\left(ucos\theta \hat{i}+(usin\theta -gt)\hat{j}\right)=0\phantom{\rule{0ex}{0ex}}{u}^{2}{\mathrm{cos}}^{2}\theta +{u}^{2}{\mathrm{sin}}^{2}\theta -\left(u\mathrm{sin}\theta \right)gt=0\phantom{\rule{0ex}{0ex}}t=\frac{u}{g\mathrm{sin}\theta}=\frac{100}{10\mathrm{sin}60}=\frac{20}{\sqrt{3}}$

Regards

**
**