1)A particle is dropped from the top of the tower of height h. At the same instant another particle is projected vertically upwards from the bottom of the tower with such a velocity that it will be able to just reach the top of the tower. When and at what height from ground the two particles meet?

Question

Sumit Majumdar · Answer

Let both the particles meet after time t.
Hence,
The distance covered by the particle coming down from the top of the tower, would be equal to:
$s=\frac{1}{2}gt^2$
The distance covered by the particle going from ground would be given by:
$h-s=ut-\frac{1}{2}gt^2$
so, the time taken by both to meet, would be given by:
$t=\frac{h}{u}$
So the height from the ground would be equal to:
$h-s=ut-\frac{1}{2}gt^2=u\times\frac{h}{u}-\frac{1}{2}g\times\left(\frac{h}{u} \right )^2=h-\frac{gh^2}{2u^2}$

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1)A particle is dropped from the top of the tower of height h. At the same instant another particle is projected vertically upwards from the bottom of the tower with such a velocity that it will be able to just reach the top of the tower. When and at what height from ground the two particles meet?

1)A particle is dropped from the top of the tower of height h. At the same instant another particle is projected vertically upwards from the bottom of the tower with such a velocity that it will be able to just reach the top of the tower. When and at what height from ground the two particles meet?

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