Question

An object $\text{A}$ is kept fixed at the point $x=3\text{m}$ and $y=1.25\text{m}$ on a plank $\text{P}$. At time $t=0$, the plank starts moving along the $+x$ - direction with an acceleration $1.5{\text{m/s}}^2$. At the same instant, a stone is project from the origin with a velocity $\overrightarrow{u}$ as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of ${45}^{\circ }$ to the horizontal. All the motions are in $x-y$ plane. The velocity $\overrightarrow{u}$ is

• A. $(6.25\hat{i}-3.75\hat{j})\text{m/s}$
• B. $(6.25\hat{i}+3.75\hat{j})\text{m/s}$
• C. $(3.75\hat{i}+6.25\hat{j})\text{m/s}$
• D. $(3.75\hat{i}-6.25\hat{j})\text{m/s}$

Answer 1

The correct option is C $(3.75\hat{i}+6.25\hat{j})\text{m/s}$

Let us suppose $t$ be the time after which the stone hits the object and $\theta$ is the angle of projection.


From the data given in the question, we have $y=1.25\text{m}.$

For vertical motion of the stone,
Using kinematic relation , $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ we get ,

$1.25=\left(u\sin \theta \right)t-\frac{1}{2}g{t}^2$

$\Rightarrow \left(u\sin \theta \right)t=1.25+5t^2........\left(1\right)$

From the data given in the question $x=3\text{m},a_x=1.5{\text{m/s}}^2$ and $u_x=0$ for the object and plank at $t=0$

For horizontal motion of the object and plank
$x=3+u_{x}t+\frac{1}{2}{a}_x{t}^2$

$x=3+0.75t^2$

So, for stone

$\left(u\cos \theta \right)t=3+0.75t^2......\left(2\right)$

Now,
Horizontal component of velocity of stone = vertical downward component [because velocity vector is inclined at ${45}^{\circ }$ with horizontal ]

Hence,
$\frac{u_y-gt}{u_x}=\tan (-{45}^{\circ })=-1$

$\therefore u\cos \theta =-(u\sin \theta -gt)$
or
$u\cos \theta =gt-u\sin \theta .........\left(3\right)$

on multiplying equation $\left(3\right)$ with $t$

$u\cos \theta t=g{t}^2-u\sin \theta t----\left(4\right)$

From equations $\left(4\right),\left(2\right)$ and $\left(1\right)$ we get,

$4.25t^2-4.25=0\Rightarrow t=1\text{s}$

Substituting $t=1\text{s}$ in $\left(1\right)$ and $\left(2\right)$, we get

$u_y=u\sin \theta =6.25\text{m/s}$

$u_x=u\cos \theta =3.75\text{m/s}$

Hence, velocity of projection of stone is
$\overrightarrow{u}=(3.75\hat{i}+6.25\hat{j})\text{m/s}$

Hence, option (c) is the correct answer.