Answer to Question #125406 in Classical Mechanics for Shinjini

Question #125406

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


1
Expert's answer
2020-07-06T17:15:23-0400

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}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS