Hitting a Bird with Projectile...

Let horizontal velocity of stone be $v_{x}$ , vertical velocity be $v_{y}$ and velocity of bird be $v_{b}$ .

Since the maximum height of the stone is $2H$ ,

$(v_{y})^{2} = (2H)(2g)$

Now, Writing equation of motion for the vertical,

$H = v_{y}t - \frac{1}{2}gt^{2}$

Let $t_{1}$ and $t_{2}$ be the times at which the stone achieves the height H. Then, $t_{1}$ and $t_{2}$ are the roots of the above equation, with $t_{1}$ > $t_{2}$

$t_{1} = \dfrac{v_{y} + \sqrt{(v_{y})^{2} - 2Hg}}{g} and~~ t_{2} = \dfrac{v_{y} - \sqrt{(v_{y})^{2} - 2Hg}}{g}$

Substitute value of $v_{y}$ here.

We get,

$t_{1} = \dfrac{ 2\sqrt{Hg} + \sqrt{2Hg}}{g} and~~ t_{2} = \dfrac{ 2\sqrt{Hg} - \sqrt{2Hg}}{g}$

$t_{1} - t_{2} = 2\dfrac{\sqrt{2Hg}}{g}$

NOTE that if the stone is to hit the bird, then they must cover same distances.

$v_{x} ( t_{1} - t_{2} ) = v_{b} t_{1}$

Substituting,

$2~v_{x}\sqrt{2Hg} = v_{b}( 2\sqrt{Hg} + \sqrt{2Hg} )$

$\dfrac{v_{b}}{v_{x}} = \dfrac{2}{\sqrt{2} + 1}$

Thus, $a + b + c + d = 6$

This problem deals with a parabola and it's generic. Speeds are proportional to distances Imgur This is a parabola with equation $y=x^{2}$ . Now, Co-ordinates of the point shown are $(-1,1), (1,1) and (\sqrt2,2)$ . Thus the required ratio is $\frac{2}{1+\sqrt2}$ . Hence the answer $\boxed{6}$

Imgur

Let the time for the STONE to rise from ground to $\displaystyle 2H$ and fall from $\displaystyle 2H$ to ground each be equal to $\displaystyle T$ .

$Let~the~time~for~the~STONE~to~rise~from~H~to~2H~and\\ fall~from~2H~to~H~each~be~equal~to~~t.\\ Total~ horizontal~distance~and~velocity~of~the~bird~are~say~\\ D_B~and~V_B; ~and~those~of~the~stone,~ D_S~and~V_S.\\ H=(1/2)gt^2~and~~~2H=(1/2)gT^2.~~~~~\therefore T=\sqrt2 *t.$ ~~~\color \red{The~ time~ taken~ by~each~ is~ the~ same}=Total~ time =TT. $\therefore TT=T+t=\sqrt2 *t+t = (1+\sqrt2)*t \\$
$Equal~Time~METHOD:-\\ D_B=V_S*(t+t),~~D_S=V_S*TT ~~~~$ \displaystyle \therefore \dfrac{V_B}{V_S}= \dfrac {D_B/TT}{D_S/TT}=\dfrac{V_S * (2t)}{V_S*TT}= \frac{2t}{(1*\sqrt2+1)*t}~~=\frac{a}{b * \sqrt c+d}. \\a+b+c+d= 2+1+2+1=\boxed{6}\\\\~~\\~~\\OR\\ \\\color \red{Equal~Distance~METHOD:-} \\The~bird~covers~D_B~with~V_B~in~Time~TT=(1+\sqrt2)*t. \\Same~distance~is~covered~by~the~stone~~with~V_S~~in~2t.\\ \displaystyle \therefore~V_B*(\sqrt2+1)*t=V_S*2*t...\displaystyle {.....\frac{V_B }{V_S }=\frac{2t }{(1*\sqrt2+1)*t}=\frac{a}{b*\sqrt c+d } }\\a+b+c+d =2+1+2+1 =6.\\\boxed{6}

We know that a projectile is a parabola. Consider the topmost point as the origin (O). Let bird be at A initially and let projectile be fired from P. Along Y-axis, OA is equal to H and OP is 2H.

By nature of parabola, along X-axis if OA is taken as 1 (say) then OP will be sqrt(2). As time required by bullet and bird is same, speeds are proportional to distance.

Horizontal distance by bird = 1 + 1 = 2 Horizontal distance by bullet = sqrt(2) + 1 which gives us a=2, b=1, c=2, d=1 and answer = 6.