An-object-A-is-kept-fixed-at-the-point-x-3-m-and-y-1-25-m-on-a-plank-P-raised-above-the-ground-At-time-t-0-the-plank-starts-moving-along-the-x-direction-with-an-acceleration-1-5-ms-2-At-

Question Number 18142 by Tinkutara last updated on 15/Jul/17

An object A is kept fixed at the point

x = 3 m and y = 1.25 m on a plank P

raised above the ground . At time t = 0,

the plank starts moving along the x -

direction with an acceleration 1.5 {ms}^{- 2} .

At the same instant a stone is projected

from the origin with a velocity \vec{u} as

shown . A stationary person on the

ground observe the stone hitting the

object during its downward motion at

an angle of 45 ° with the horizontal .

Take g = 10 m / s^{2} and consider all

motions in the x - y plane .

1 . The time after which the stone hits

the object is

2 . The initial velocity (\vec{u}) of the

particle is

Commented by Tinkutara last updated on 15/Jul/17

Answered by ajfour last updated on 16/Jul/17

let initial position of object be

P (a, b) \Rightarrow a = 3 m, b = 1 . 25 m

let stone hits object at x = d, y = b .

let time taken = t_{0}

In this time object moves from

x = a to x = d with acceleration

A = 1 . 5 m / s^{2} . So

d = a + \frac{1}{2} {At}_{0}^{2} \dots . (i)

As stone hits object while

descending at α = 45 °, we observe

v_{y} = - u_{x}, v_{x} = u_{x} \dots . (ii)

d = u_{x} t_{0} \dots (iii)

from (i) and (iii) :

a + \frac{{At}_{0}^{2}}{2} = u_{x} t_{0} \dots (iv)

Also b = u_{y} t_{0} - \frac{{gt}_{0}^{2}}{2} \dots . (v)

And [v_{y} = - u_{x} = u_{y} - {gt}_{0}] \times t_{0}

or - u_{x} t_{0} = u_{y} t_{0} - {gt}_{0}^{2} \dots (vi)

(v) - (vi) gives

b + u_{x} t_{0} = \frac{{gt}_{0}^{2}}{2} . And from eq . (iv)

b + (a + \frac{{At}_{0}^{2}}{2}) = \frac{{gt}_{0}^{2}}{2}

substituting values

\frac{5}{4} + 3 + \frac{3}{4} t_{0}^{2} = {5 t}_{0}^{2}

{17 t}_{0}^{2} = 17

\Rightarrow t_{0}^{2} = 1

\Rightarrow t_{0} = 1 s

so u_{x} = \frac{a}{t_{0}} + \frac{{At}_{0}}{2} = 3 + \frac{3}{4} = \frac{15}{4} m / s^{2}

u_{y} = \frac{b}{t_{0}} + \frac{{gt}_{0}}{2} = \frac{5}{4} + 5 = \frac{25}{4} m / s^{2}

\vec{u} = \frac{5}{4} (3 \hat{i} + 5 \hat{j}) m / s .

Commented by ajfour last updated on 16/Jul/17

view again, it is shorter now .

Commented by Tinkutara last updated on 16/Jul/17

Now it is better .

Commented by ajfour last updated on 16/Jul/17

Is my answer correct ?

Commented by Tinkutara last updated on 16/Jul/17

Yes, Thanks Sir!