Question

A particle is projected from point $\text{O}$ on the ground with velocity $u=5\sqrt{5}\text{m/s}$ at angle $\alpha ={\tan }^{-1}\left(0.5\right)$. It strikes at point $\text{C}$ on a fixed smooth plank $\text{AB}$ having inclination ${37}^{\circ }$ with horizontal as shown in figure. If the particle does not rebound, the co-ordinate of point $\text{C}$ in reference to co-ordinate system as shown in the figure is
(Given that $OA=\frac{10}{3}\text{m}$)

• A. $(7\text{m},1.25\text{m})$
• B. $(9\text{m},1.25\text{m})$
• C. $(5\text{m},1.25\text{m})$
• D. $(8\text{m},1.25\text{m})$

Answer 1

The correct option is C $(5\text{m},1.25\text{m})$

Given, velocity of the particle at point $\text{O}$, $u=5\sqrt{5}\text{m/s}$.

For given projectile along $\text{x-axis}$,
component of speed, $u_x=5\sqrt{5}\cos \alpha$.

Substituting the value of $\cos \alpha$, in the above equation we get,
$u_x=5\sqrt{5}\times \frac{2}{\sqrt{5}}$

$\therefore u_x=10\text{m/s}$

Time taken to reach at point $\text{D}$,

$t=\frac{\text{OD}}{u_x}$

$\Rightarrow t=\frac{\frac{10}{3}+x}{10}$

$\Rightarrow t=\frac{10+3x}{30}.........\left(1\right)$

Component of velocity along $\text{y-axis}$,

$u_y=5\sqrt{5}\sin \alpha$

Substituting the value of $\sin \alpha$ from above triangle,

$\Rightarrow u_y=5\sqrt{5}\times \frac{1}{\sqrt{5}}$

$\therefore u_y=5\text{m/s}$

Applying equation of motion,$s=ut+\frac{1}{2}a{t}^2$, along $\text{CD}$,

where, $s=CD=x\tan {37}^{\circ }$,

$t=\frac{10+3x}{30}$,

$a=-g=-10\text{m}{\text{s}}^{-2}$

Substituting the values in the equation of motion,
$\Rightarrow x\tan {37}^{\circ }=5\left(\frac{10+3x}{30}\right)-\frac{1}{2}\times 10x{\left(\frac{10+3x}{30}\right)}^2$

$\Rightarrow \frac{3x}{4}=\frac{10+3x}{6}-5{\left(\frac{10+3x}{30}\right)}^2$

$\Rightarrow \frac{3x}{4}=\frac{10+3x}{6}-\frac{5\times (10+3x{)}^2}{30\times 30}$

$\Rightarrow \frac{3x}{4}=\frac{30(10+3x)-(10+3x{)}^2}{30\times 6}$

$\Rightarrow 135x=300+90x-100-9x^2-60x$

$\Rightarrow 9x^2+105x-200=0$

On solving this, we get, $x=\frac{5}{3}$ and $x=-\frac{40}{3}$

Since, distance cannot be negative.

Thus, $OD=\frac{10}{3}+x=\frac{10}{3}+\frac{5}{3}=5\text{m}$

and $CD=\frac{3x}{4}=\frac{3}{4}\times \frac{5}{3}=\frac{5}{4}=1.25\text{m}$

Hence, co-ordinate of point $\text{C}$ is $(5\text{m},1.25\text{m})$