Question

a particle is released from rest from a tower of height 3h .The ratio of times to fall equal height h i.e ,t1 : t2 : t3​

Answer 1

ANSWER:

Given:

• A particle is released from a tower of height 3h.

To Find:

• Ratio of time intervals to cover equal height 'h' [t$_1$ : t$_2$ : t$_3$ ]

Diagram:

Refer attachment

Solution:

• Time taken from point A to point B = t$_1$
• Time taken from point B to point C = t$_2$
• Time taken from point C to point D = t$_3$

As the particle is released, the initial velocity (u) = 0

The acceleration is g(due to gravity).

We know that,

$s=ut+^1\!\!\!/_2at^2 -------(\text{2 ^{nd} equation of motion})$

So,

1) For particle traveling from point A to point B:

⇒ s = h, t = t$_1$

$:\implies h = (0)(t_1) + ^1\!\!\!/_2(g)(t_1)^2 \\\\:\implies h = ^1\!\!\!/_2gt_1^2 \\\\:\implies 2h = gt_1^2 \\\\:\implies t_1^2 = \frac{2h}{g} \\\\:\implies t_1 = \sqrt{\frac{2h}{g}} -----(1)$

2) For particle traveling from point A to point C:

⇒ s = 2h, t = t$_1$ + t$_2$

$:\implies 2h = (0)(t_1+t_2) + ^1\!\!\!/_2(g)(t_1+t_2)^2 \\\\:\implies 2h = ^1\!\!\!/_2g(t_1+t_2)^2 \\\\:\implies 4h = g(t_1+t_2)^2 \\\\:\implies (t_1+t_2)^2 = \frac{4h}{g} \\\\:\implies t_1 +t_2= \sqrt{\frac{4h}{g}} -----(2)$

3) For particle traveling from point A to point D:

⇒ s = 3h, t = t$_1$ + t$_2$ +t$_3$

$:\implies 3h = (0)(t_1+t_2+t_3) + ^1\!\!\!/_2(g)(t_1+t_2+t_3)^2 \\\\:\implies 3h = ^1\!\!\!/_2g(t_1+t_2+t_3)^2 \\\\:\implies 6h = g(t_1+t_2+t_3)^2 \\\\:\implies (t_1+t_2+t_3)^2 = \frac{6h}{g} \\\\:\implies t_1 +t_2+t_3 = \sqrt{\frac{6h}{g}} -----(3)$

Now,

Time taken from B to C (t$_2$)

⇒ Time taken from A to C - Time taken from A to B

From (1) & (2)

$:\implies \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} \\\\:\implies \left(\sqrt{\frac{2h}{g}}\right)*(\sqrt2-1) -------(4)$

Time taken from C to D (t$_3$)

⇒ Time taken from A to D - Time taken from A to C

From (2) & (3)

$:\implies \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} \\\\:\implies \left(\sqrt{\frac{2h}{g}}\right)*(\sqrt3-\sqrt2) -------(5)$

Therefore,

$:\implies t_1 : t_2 : t_3 \\\\\text{From (1), (4) & (5)} \\\\:\implies \sqrt{\frac{2h}{g}} : \left(\sqrt{\frac{2h}{g}}\right)*(\sqrt2-1) : \left(\sqrt{\frac{2h}{g}}\right)*(\sqrt3-\sqrt2) \\\\\text{But \left(\sqrt{\frac{2h}{g}}\right) gets cancelled, }\\\\\bf{:\implies 1 : \sqrt2 - 1 : \sqrt3 - \sqrt2}$

Formula Used:

• s = ut + ¹/₂ at²