Question

A particle is released from rest from a tower of height 3h.the ratio of times to fall equal heights h ie t1:t2:t3 is

Answer 1

Answer:

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Answer 2

Answer:

The ratio of times to fall equal heights h is 1 : (√2-1) :(√3-√2).

Explanation:

The particle covers the equal height h from point A to B in time t₁ , A to C in time t₂ and A to D in time t₃.

From the height A to B:-

From the 2nd equation of motion:

           $h=(0)(t_{1})+\frac{1}{2}gt^{2}_{1} \\ t_{1}=\sqrt{\frac{2h}{g} }$               .............(1)

From height A to C:-

            $2h=(0)(t_{2})+\frac{1}{2}gt^{2}_{2} \\ t_{2}=\sqrt{\frac{4h}{g} }$             ...........(2)

From point A to D:-

            $3h=(0)(t_{3})+\frac{1}{2}gt^{2}_{3} \\ t_{3}=\sqrt{\frac{6h}{g} }$            ...............(3)

From  A to B time taken   $t_{1}=\sqrt{\frac{2h}{g} }$

From  B to C time taken  $t_{2}-t_{1}=\sqrt{\frac{4h}{g} }-\sqrt{\frac{2h}{g} }=\sqrt{\frac{2h}{g} }(\sqrt{2} -1)$

From  C to D time taken  $t_{3}-t_{2}=\sqrt{\frac{6h}{g} }-\sqrt{\frac{4h}{g} }=\sqrt{\frac{2h}{g} }(\sqrt{3} -\sqrt{2} )$

Therefore the ratio  $= 1:(\sqrt{2} -1):(\sqrt{3} -\sqrt{2} )$