Projectile over a Square (part -2)

Let AC be the y-axis and the horizontal be the x-axis. It can easily be seen that the equation of parabola passing through points B,D and C is $-10\sqrt2 (y-20\sqrt2) = x^{2}$ by basic transformations of $y=x^{2}$ . Find the slope of this by differentiating with respect to x at x=-20; we get $\frac { dy }{ dx } _{ x=-20 }=\frac { -2x }{ 10\sqrt { 2 } } =2\sqrt { 2 }$ . Hence the answer $\boxed{4}$

Let velocity of the particle at the time of projection be $u$ & at point B be $v$ .

First let concentrate on part BCD of the Projectile. It can also be called as a projectile with initial velocity $u$ , at an angle $\alpha$ from horizontal, $H_{max} = H$ & Range $(R_{1})$ = $2H$ .

So, $tan\alpha = \frac{4H_{max}}{R_{1}} = 2$

And $H = \frac{{u}^{2}{sin}^{2}\alpha}{2g}$

$\implies$ $u sin\alpha = \sqrt{2gH}$

$\implies$ $u cos\alpha = \sqrt{2gH}\times cot\alpha$

or, $u cos\alpha = \boxed{\sqrt{\frac{gH}{2}}}$ .

Now come to the projectile from ground.

$\frac{{v}^{2}{sin}^{2}\theta}{2g} = 2H$

$\implies$ $v sin\alpha = \sqrt{4gH}$

$\implies$ $v cos\theta = \boxed{\sqrt{4gh}\times cot\theta}$ .

Since $v cos\theta = u cos\alpha$ so $\sqrt{4gh}\times cot\theta = \sqrt{\frac{gH}{2}}$ .

Therefore $\boxed{tan\theta = 2\sqrt{2}}$

Hence the answer is $2+2=4$ .