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Answers
Here's the Answer ❗
→ 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
Put the values here ,
From here time t
Now by using the first equation of motion
v = u + at
[ 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 ,
[ 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
Now here v = u - gt
Va = 2 ( Vb)
As per the question we have
•°• ( 1 - x ) = ±⅓
So we have ,
•°• →
Answer:
please refer to the attachment
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thank you.
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anyways,
u just study well.
practice more.
forget less.
think most.
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