Answer to Question #125406 in Classical Mechanics for Shinjini

Question

when a stone has fallen a distance x after being dropped from the top of a tower another stone is dropped from point y below the top op the tower, if both reach the ground at the same time prove that height of the tower is (x+y)²/4x

Answer 1

As per the given question,

Let the time taken to reach the second stone to reach the bottom is t,

Initial velocity of the stone x and stone is =0 and for stone y is also u=0

As per the given question,

Let the height of the tower is h, and let time taken by the first stone to reach the bottom of the tower $t_h$ , let the time taken by the stone x to reach the x distance is $t_x$ and the time taken by the stone y to reach to the bottom of the tower is $t_y$,

Hence,

$x=ut+\frac{gt_x^2}{2}$

$\Rightarrow x =0+\frac{gt_x^2}{2}$

$\Rightarrow t_x =\sqrt{\frac{2x}{g}}$

time taken by the stone x to reach to the bottom of the tower is $h=ut_h+\frac{gt_h^2}{2}$

$\Rightarrow h=0+\frac{gt^2}{2}$

$\Rightarrow t_h=\sqrt{\frac{2h}{g}}$

time taken by stone y to reach the bottom of the tower is $h-y=ut_y+\frac{gt_y^2}{2}$

$\Rightarrow h-y=0+\frac{gt_y^2}{2}$

$\Rightarrow t_y =\sqrt{\frac{2(h-y)}{g}}$

Now,

$t_y=t_h-t_x$

Now, substituting the values,

$\Rightarrow \sqrt{\frac{2(h-y)}{g}}=\sqrt{\frac{2h}{g}}-\sqrt{\frac{2x}{g}}$

$\Rightarrow \sqrt{2(h-y)}=\sqrt{2h}-\sqrt{2x}$

$\Rightarrow \sqrt{x}=\sqrt{h}-\sqrt{h-y}$

now, squaring both side,

$x=h+(h-y)-2\sqrt{h(h-y)}$

$\Rightarrow x=2h-y-2\sqrt{h(h-y)}$

$\Rightarrow 2h-2\sqrt{h(h-y)}=x+y$

Now,

$\Rightarrow 2h-(x+y)=2\sqrt{h(h-y)}$

squaring both side,

$\Rightarrow (2h-(x+y))^2=4(h^2-hy)$

$\Rightarrow 4h^2+(x+y)^2-4h(x+y)=4h^2-4hy$

$\Rightarrow (x+y)^2-4hx-4hy=-4hy$

$\Rightarrow 4hx=\frac{(x+y)^2}{}$

$\Rightarrow h=\frac{(x+y)^2}{4x}$