One Bird With The Same Stone... Twice!

Point A is point of projection of stone

Point B is top of the tree of height 'h'

Point C is maximum height of stone

D is point where stone hit the bird

E is point of landing of stone

Let time from A to C is t , similarly time from C to E is also t

Let time from B to C is t' , similarly time from C to D is also t'

At point C , H = 0.5g(t')² , 2H = 0.5gt²

By this t=√2t'

When stone reach at point D , time of bird from B to D is t+t'

Let speed of bird is V

V(t+t')=Ux(2t')

V(√2t'+t')=Ux(2t')

(√2+1)\2 = (Ux)/V

1/√2 + 1/2 = Ux/V

So n = 2 and 2n = 4

Great question!

Let us suppose the height of the tree to be $\large h$ .
Then, the maximum height attained by the ball is $\large 2h$ .

The premise of the question is that the ball is thrown with some initial velocity $\large v$ . The bird starts to fly horizontally at the same instant with some velocity $\large b$ . This implies it stays at the same height $\large h$ from the ground. The ball goes to a height $\large 2h$ , starts to descend and then hits the bird.

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1. Firstly, lets find the initial vertical velocity $\large v_{y}$ of ball

Maximum height attained by the ball will be $\large \frac{v_{y}^2}{2g}$ , which should be equal to 2h. Hence, $\large v_{y}$ = $\large 2\sqrt{hg}$

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1. Second, we find the times at which the ball attains the height h .

Writing our equation for block's vertical motion yields

$\large s_{y}$ = $\large u_{y}t$ - $\large \frac{g}{2t^2}$ .

The quadratic above gives two values of t, $\large \frac{2\sqrt{hg} - \sqrt{2hg}}{g}$ & $\large \frac{2\sqrt{hg} + \sqrt{2hg}}{g}$ .

Clearly, the former is the time when ball attains h height while going up and latter is the time for which it is at the same height h while descending down.

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1. Let's find the answer!

The bird flies for a duration of $\large \frac{2\sqrt{hg} + \sqrt{2hg}}{g}$ at velocity b .

This distance would be the same as the difference in the horizontal distances covered by the stone in between the two times i.e. $\large (\frac{2\sqrt{hg} + \sqrt{2hg}}{g} - \frac{2\sqrt{hg} - \sqrt{2hg}}{g})u_{x}$

Equate and solve to find $\large \frac{u_{x}}{b} = \frac{1}{\sqrt{2}} + \frac{1}{2}$

posted a pic of the solution as i found it cumbersome to type all those terms

Great use of projectile. I have done it using the time taken by the stone to reach height H/2 secomd time is equal to time taken by bird to reach same position from the first position.