Q: 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 u as shown in below figure.

A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45° to the horizontal. At the motions are in xy-plane. Find u and the time after which the stone hits the object. (Given, g = 10 ms^{-2})

**Solution :** Let t be the time after which the stone hits the object and θ be the angle which the velocity vector u makes with horizontal . According to question , we have following three conditions .

Vertical displacement of stone is 1.25 m

1.25 = (u sinθ ) t – (1/2)g t^{2}

1.25 = (u sinθ ) t – 5 t^{2}

(u sinθ ) t = 1.25 + 5 t^{2} …(i)

Horizontal displacement of stone = 3 + displacement of object A

(u cosθ ) t = 3 + (1/2)a t^{2} ; Where a = 1.5 m/s^{2}

(u cosθ ) t = 3 + 0.75 t^{2} …(ii)

Horizontal component of velocity (of Stone) = Vertical component

(because velocity vector is inclined at 45° with horizontal )

u cosθ = g t – u sinθ …(iii)

(As stone is in its downward direction)

Therefore g t > u sinθ . In upward motion u sinθ > g t

Multiplying Equation (iii) with t ,

u cosθ t + u sinθ t = g t^{2}

u cosθ t + u sinθ t = 10 t^{2} …(iv)

Now Equation (iv) , (ii) & (i) gives

4.25 t^{2} – 4.25 = 0

t= 1 sec

putting t = 1 sec in Equation (i) & (ii) we get

u sinθ = 6.25 m/s or , u_{y} = 6.25 m/s

u cosθ = 3.75 m/s or , u_{x} = 3.75 m/s

$\displaystyle \vec{u} = u_x \hat{i} + u_y \hat{j}$

$\displaystyle \vec{u} = 3.75 \hat{i} + 6.25 \hat{j}$ m/s