Question

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 $1.5 m/s^2$. At the same instant a stone is projected from the origin with a velocity $\vec{\upsilon}$ 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. Find $\vec{\upsilon}$ and the time after which the stone hits the object.$[Take g = 10 m/s^2]$

Answer 1

Answer:

Velocity 7.2m/s and time 1 seconds.

Explanation:

If we let the angle of projection to be x so the components will be vx = ucosx and vy=usinx.

Now, from the questions data we have ucosx=vcos45 or v=u√2cosx.

vy=usinx - gt = -vsin45.

usinx - gt=-ucosx.

usinx=gt-ucosx.

Now, from the main equation of

y=u(sinx)t - 1/2gt^2.

We can substitute that :-

gt^2 - utcosx -gt^2/2=1.25.

Again, we have utcosx=3 + 3/4gt^2.

Now, from both the equations after adding them we will get the value of time to be 1seconds.

Again, we have usinx = 5 + 1.25=6.25.

ucosx=3.75.

On solving we get the value of u to be 7.2.

So, the value of velocity is 7.2m/s.