Question

Two bodies of masses m1
and m2
are let fall freely from heights 1
h and 2
h respectively. The ratio of
time taken by the bodies to fall through these heights is

Answer 1

Question

Two bodies of masses m1 and m2 are fallen freely from heights h1 and h2 respectively. The ratio of time taken by the bodies to fall through these heights is.

Solution

For first body:

Mass (m) = m1

Height (h) = h1

For second body:

Mass (m) = m2

Height (h) = h2

Given that, Two bodies of masses m1

and m2 are fallen freely from heights h1 and h2 respectively.

For the freely falling body,

a = g = 9.8 m/s² or 10 m/s² (approx.)

Initial velocity (u) = 0 m/s

Using Second Equation of Motion i.e.

s= ut + 1/2 at²

Assume that, the first body takes t1 time and second body takes t2 time.

For the first body:

h1 = 0(t1) + 1/2 g(t1)²

h1 = 1/2 g(t1)²

2(h1) = g(t1)²

[2(h1)]/g = (t1)²

t1 = √[2(h1)/g]

For the second body:

h2 = 1/2 g(t2)²

t2 = √[2(h2)/g]

We have to find the ratio of time taken by the bodies to fall through these heights.

→ t1/t2 = [√2h1/g] / [√2h2/g]

→ t1/t2 = √h1/√h2

Answer 2

Correct Question:

Two bodies of masses m₁  and m₂  are let fall freely from heights H₁ and H₂ respectively. The ratio of  time taken by the bodies to fall through these heights is?

Answer:

• The ratio of time period is √H₁ : √H₂

Given:

1. Heights through which they fall = h₁ and h₂
2. As they fall, the initial velocity will be zero.

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bullet\;\underline{\boxed{\sf H=u\;t+\dfrac{1}{2}\;a\;t^{\;2}}}$

Here,

• H Denotes Height.
• u Denotes Initial velocity.
• t Denotes time taken
• a Denotes acceleration.

Now,

$\displaystyle\Rightarrow\sf H=u\;t+\dfrac{1}{2}\;a\;t^{\;2}$

As the ball is falling freely the initial velocity will be zero.

$\displaystyle\Rightarrow\sf H=0\times t+\dfrac{1}{2}\;a\;t^{\;2}\\\\\\\Rightarrow\sf H=\dfrac{1}{2}\;a\;t^{\;2}$

From the above conclusion we can get,

$\displaystyle\Rightarrow\sf H\propto t^{\;2}\\\\\\\Rightarrow\sf \sqrt{H} \propto t\\\\\\\Rightarrow\sf t \propto \sqrt{H}$

Now, Substituting

$\displaystyle\Rightarrow\sf \dfrac{t_1}{t_2}=\sqrt{\dfrac{H_a}{H_b}}$

we Know,

• H_a = H₁
• H_b = H₂

Substituting and solving,

$\displaystyle\Rightarrow\sf \dfrac{t_1}{t_2}=\sqrt{\dfrac{H_1}{H_1}}\\\\\\\Rightarrow\sf t_1\;\colon t_2=\sqrt{H_1}\;\colon\sqrt{H_2}\\\\\\\Rightarrow\sf \large{\underline{\boxed{\red{\sf t_1\;\colon t_2=\sqrt{H_1}\;\colon\sqrt{H_2}}}}}$

∴ The ratio of time period is √H₁ : √H₂.

Note:

• H_a is the height achieved by m₁ mass.
• H_b is the height achieved by m₂ mass.

$\rule{300}{1.5}$