# Question #f4ddf

Mar 18, 2017

Given that the initial(at $t = 0$) velocity of projection of the particle is $u = 40 \text{m/s}$ in the direction making an angle $\theta = {53}^{\circ}$ (upward) with the horizontal.

Initially at position O
$t = 0$

The horizontal component of velocity ${u}_{x} = u \cos \theta$. This will remain unaltered throughout its movement as it is orthogonal to gravitational force.

The vertical component of velocity ${u}_{y} = u \sin \theta$ . This will change as the force of gravitation acts opposite to it. .

Let at position P the velocity of the particle becomes perpendicular to the initial velocity $u$. So at this position the velocity of the particle will subtend angle $= {90}^{\circ} - \theta$ (downward) with the horizontal

At position P

The horizontal component of velocity ${u}_{x} = u \cos \theta$.
The actual velocity $v$ at this position is inclined with the constant horizontal component at an angle $90 - \theta$.

So
$\implies v \cos \left(90 - \theta\right) = u \cos \theta$

$\implies v \sin \theta = u \cos \theta$

$\implies v = \frac{u \cos \theta}{\sin} \theta = u \cot \theta$

Now acceleration due to gravity $g$ is always orthogonal to horizontal component of velocity but it is inclined with the velocity $v$ at position P at an angle $\theta$ . So the component of acceleration due to gravity orthogonal to v is $a = g \sin \theta$

This acceleration (a) being orthogonal with v can be treated as centripetal acceleration and its relation with radius of curvature $r$ is as follows

$a = {v}^{2} / r$

$\implies g \sin \theta = \frac{{u}^{2} {\cot}^{2} \theta}{r}$

$\implies r = \frac{{u}^{2} {\cot}^{2} \theta}{g \sin \theta}$

Inserting $u = 40 \text{m/s"; theta =53^@;g=10"m/} {s}^{2}$
$\cot {53}^{\circ} \approx 0.75 \mathmr{and} \sin {53}^{\circ} \approx 0.8$

we get

$r = \frac{{40}^{2} {\cot}^{2} 53}{10 \times \sin 53} = 112.5 m$ which is option (3)