Question

A body is dropping from the top of the tower of height h.At the same time another body is thrown vertically upward such that it just approaches to the top of tower. Where and when the two bodies crosses each other.
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Answer 1

Explanation:

body will meet in 3h/4 m above the ground

Answer 2

This type of question, you can easily solve by using relative concept.

We should use formula,

time taken to cross each other = relative seperation/relative velocity.

here, A body is dropped from a tower of height h.

so, initial velocity of body is 0 m/s

another body is thrown vertically upward such that it just approaches to the top of tower.

Let initial velocity of 2nd body is u.

now using formula, v² = u² + 2as

at highest position, velocity of 2nd body equals zero. so, 0 = u² + 2(-g)h

or, u = √(2gh)

now, time taken to cross each other = (h-0)/{0 + √(2gh)}

= h/√(2gh) = √{h/2g}

now, position where they cross each other, s =ut - 1/2gt²

= √(2gh) × √{h/2g} - 1/2 × g × h/2g

= h - h/4

= 3h/4

hence, at 3h/4 from the ground they cross each other and time taken to cross each other is √{h/2g}