Ball A is dropped from the top of a building and at the same instant that a ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions and the speed of A is twice the speed of B. At what fraction of the height of the building did the collisionoccur?

Ball A is dropped from the top of a building and at the same instant t

1.	Ball A is dropped from the top of a building and at the same instant that a ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions and the speed of A is twice the speed of B. At what fraction of the height of the building did the collisionoccur?
Answer» <html><body><p></p>Solution :Given `V_(A)=2V_(B)` <br/> `impliesg t=2(u-<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> t)impliest=(2u)/(3g)` <br/> Displacement of A is `S_(A)=(1)/(2)g t^(2)=(2u^(2))/(9g)` <br/> Displacement of B is `S_(B)=ut-(1)/(2)g t^(2)=(4u^(2))/(9g)` <br/> Now fraction of hight of the <a href="https://interviewquestions.tuteehub.com/tag/building-905221" style="font-weight:bold;" target="_blank" title="Click to know more about BUILDING">BUILDING</a> where collision <a href="https://interviewquestions.tuteehub.com/tag/occur-583012" style="font-weight:bold;" target="_blank" title="Click to know more about OCCUR">OCCUR</a> is `(h_(B))/(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)=((4u^(2))/(9g))/((4u^(2))/(9g)+(2u^(2))/(9g))=(2)/(3)`</body></html>