Question

A body is allowed to fall from a height of 100 m. If time taken for first 50 m is t1 and for remaining 50 m is t2, then ratio of t1 & t2 is
(1) 5 : 2
(2) 2 : 5
(3) 3 : 4
(4) 4 : 3

Answer 1

Time taken for 1st 50 m
t1 = √(2H/g)
= √(2 × 50 / 10)
= √10

Time taken fall 100 m
T = √(2H / g)
= √(2 × 100 / 10)
= √20

Time taken to fall last 50 m
t2 = T - t1
= √20 - √10
= √10 (√2 - 1)

t1 / t2 = √10 / [√10 (√2 - 1))]
= 1 / (√2 - 1)
= 2.41
≈ 5 / 2

Ratio is approximately 5 : 2
[Note : I took g = 10 m/s^2 Maybe we will get exactly 5 : 2 if we take g = 9.8 m/s^2]

Answer 2

Concept

The second law on motion is $v=ut+\frac{1}{2}at^{2}$

where u is initial velocity, t is time, v is final velocity and a is acceleration.

Given

A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is $t_1$ and for the remaining 50 m is $t_{2}$.

To find

We have to find the ratio of $t_{1}$ and $t_{2}$.

Solution

Here a=g (gravity), u= 0 (At starting) and s= 100 (height),

Time taken for 1st 50 m

$t_1 = \sqrt{\frac{2H}{g} } \\= \sqrt{\frac{2(50)}{(10)} }\\= \sqrt{10}$

Time is taken fall 100 m

$T= \sqrt{\frac{2H}{g} } \\= \sqrt{\frac{2(100)}{(10)} }\\= \sqrt{20}$

Time taken to fall last 50 m

$t_2=T-t_1\\=\sqrt{20}-\sqrt{10} \\=\sqrt{10}(\sqrt{2}-1)$

then

$\frac{t_1}{t_2}=\frac{ \sqrt{10}}{ \sqrt{10}(\sqrt{2}-1)}\\=\frac{1}{(\sqrt{2}-1} \\=2.41\\=\frac{5}{2}$

Hence option (1) 5 : 2 is correct.