Question

An object $A$ is kept fixed at the point $x=3m$ and $y=1.25m$ on a plank $P$ raised above the ground. At time $t=$ the plank starts moving along the positive $x$-direction with an acceleration $1.5m/s^2$. At the same instant a stone is projected from the origin with a velocity $\overrightarrow{u}$ as shown in the figure. A stationary person on the ground observes the stone hitting the object during which it makes a downwards motion at the angle of ${45}^o$ to the horizontal. All the motions are in x-y plane. Find $\overrightarrow{u}$ and the time after which the stone hits the object (Take $g=10m/s^2$)

Answer 1

Let ball hit the plank at point P then at point A angle of stone with horizontal is also $45°$.

$\left(1\right)$ x-co-ordinate of stone and plank is same.

$x=u_{x}t=3+\frac{1}{2}a{t}^2=u_{x}t=3+\frac{1}{2}\times 1.5t^2⟶1$

Since y-coordinate is same

$y=u_{y}t-\frac{1}{2}g{t}^2=1.25=\frac{5}{4}⟶2$

at point P velocity make angle $45°$. clockwise

Hence

$\frac{(u_y-gt)}{u_x}=-\tan 45°=-1u_x=-(u_y-gt)⟶3$

Substituting value of $u_{x}t$ in equation $3$ by multiplying t.

${-u}_{y}t+10t^2=3+\frac{3}{4}{t}^2{-u}_{y}t+\frac{37}{4}{t}^2=3⟶4$

equation $\left(2\right)$ can be written as $u_{y}t-5t^2=\frac{5}{4}⟶5$

adding equation $\left(5\right)$ to $\left(1\right)$ we have,

$\frac{17}{4}{t}^2=\frac{17}{4}t=1sec$

therefore equation $\left(2\right)$ abve gives

$u_y=\frac{25}{4}$ and equation $\left(3\right)$ gives

$u_x=gt-u_y=\frac{15}{4}m/s$

hence $u=(\frac{15}{4}i+\frac{25}{4}j)m/s$