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A ball is dropped from rest at height h. Directly below on the ground, a second ball is simultaneously thrown vertically upward with speed v0. If two balls collide at the moment the second ball is instantaneously at rest, the height of collision is1) h/22) h/33) h/44) 3h/4
In such questions think separately abiut ywo balls,now first of all think of the ball that is been thrown up,through kinematic eq.{v^2=u^2+2as} we can find the distance at which the ball is momentarily at rest thus v=0,tgis comes out to be,v°^2/20.Now find the time taken by the ball to reach this height with the help of kinematic eq. Then take this time and see how much distance is covered by the ball thrown from above in this mich time,it comes out to be s=v°^2/2.Now as we know that they collide at time "t" the total distace will be s+s=h thus answer is h/2
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