Question

An object is projected with speed $50\text{m/s}$ at an angle ${53}^{\circ }$ with horizontal from ground. The radius of curvature of its trajectory at $t=1\text{sec}$ after projection will be:
$($Take $g=10{\text{m/s}}^2)$

• A. $90\sqrt{2}\text{m}$
• B. $180\sqrt{2}\text{m}$
• C. $90\text{m}$
• D. $125\text{m}$

Answer 1

The correct option is B $180\sqrt{2}\text{m}$

Given:
$u=50\text{m/s};\theta ={53}^{\circ }$


At $t=1\text{sec}$ after projection,

$v_x=u_x=u\cos {53}^{\circ }$

$\Rightarrow v_x=50\times \frac{3}{5}=30\text{m/s}$

Now using equation of motion,

$v_y=u_y-a_{y}t$

$v_y=u\sin {53}^{\circ }-g\left(1\right)$

$\Rightarrow v_y=50\times \frac{4}{5}-10\left(1\right)=30\text{m/s}$

Thus, the angle made by velocity $\left(v\right)$ from horizontal is

$\tan \theta =\frac{v_y}{v_v}=1$

$\therefore \theta ={45}^{\circ }$ and

$v=\sqrt{v_x^2+v_y^2}=30\sqrt{2}\text{m/s}$

Now for radius of curvature of trajectory at $\text{P}$,


Centripetal acceleration is a component of acceleration perpendicular to the velocity.

$a_c=g\cos {45}^{\circ }$

$a_c=\frac{g}{\sqrt{2}}$

Thus, $a_c=\frac{v^2}{R}$

$\Rightarrow \frac{g}{\sqrt{2}}=\frac{(30\sqrt{2}{)}^2}{R}$

$\therefore R=\frac{\sqrt{2}\times 900\times 2}{10}=180\sqrt{2}\text{m}$

Hence, option $\left(b\right)$ is correct.

Why this question ? Tip: always remember that whenever there is a curvature, then it will have centripetal acceleration at that point directed along centre of circle, of which curvature is a part. Thus, $a_{\perp }=\frac{v^2}{R}$