Physics, asked by Anonymous, 11 months ago

please solve this question I'll mark u as BRAINLIEST no spamming​

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Answered by TheInsaneGirl
6

{\bold{\underline{Heya \:Mate!!}}}

Here's the Answer

{\bold{Question }} → A ball a dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground . When the balls collide , the speed of A is twice that of B. At what fraction of the height of the building did the collision occur?

Answer .

Firstly for ball A we have ,

Initial Velocity u = 0

Distance s = ( 1 - x )h

Using the second equation of motion

s = ut +  \frac{1}{2} at {}^{2}

Put the values here ,

 =  > (1 - x)h =  \frac{1}{2} gt {}^{2}

From here time t

t =  \sqrt{2(1 - x) \frac{h}{g} }

Now by using the first equation of motion

v = u + at

v = g \sqrt{2(1 - x) \frac{h}{g} } ..............(1)

[ as here u is zero. Instead of t putting the value we derived above ]

_________________

Now for Ball B ,

Distance travelled s = xh

Again following the steps as done for A ,

xh = u \sqrt{2(1 - x) \frac{h}{g} }  -  \frac{1}{2} g(2(1 - x) \frac{h}{g} )

[ Here again instead of t we have put the value derived as time for both is same. Now as the ball is thrown against gravity , g will be negative ]

From this equation We derive the value of u

 =  > u =   \sqrt{ \frac{gh}{2(1 - x)} }

Now here v = u - gt

 =  > v =  \sqrt{ \frac{gh}{2(1 - x)} }  -  \sqrt{2g(1 - x)h} ...........(2)

Va = 2 ( Vb)

As per the question we have

 \sqrt{2g(1 - x) h }  = 2\times  \sqrt{ \frac{gh}{2(1 - x)} }  - 2 \sqrt{2g(1 - x) h}  \\  \\  =  >3  \sqrt{} 2g(1 - x)h = 2 \sqrt{ \frac{gh}{2(1 - x)} }  \\  =  > 9(1 - x) =  \frac{1}{1 - x}  \\  \\  =  > 9(1 - x) {}^{2}  = 1 \\  =  > (1 - x) {}^{2}  =  \frac{1}{9}  \\  \\

•°• ( 1 - x ) = ±⅓

So we have ,

{ \bold{{taking \: positive\: value \:}}} =  >  \:  \\  =  > 1 - x =   \frac{1}{3}  \\  =  > x = 1  -  \frac{1}{3}  =  \frac{2}{3}

•°• {\underline{\underline{Answer}}}\frac{2}{3}

{\boxed{\bold{Hope\:Helps!!}}}

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Anonymous: y am I unable to mark u as brainliest
Anonymous: ok gud night
Anonymous: bye
TheInsaneGirl: You can mark when someone else answers or when the option comes.
Anonymous: made it
TheInsaneGirl: :)
Anonymous: :)
Anonymous: hey i got one other question
Anonymous: please solve it
Anonymous: I have posted the question
Answered by Anonymous
18

Answer:

please refer to the attachment

I hope it would help you

thank you.

and before I would have checked my inbox, I think u again blocked.

and ur status might be for me.

good line it is.

anyways,

u just study well.

practice more.

forget less.

think most.

THANK YOU ‼️‼️✴️✴️✴️✴️

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