Question

Two bodies A (of mass 1 kg) and B (of mass
3 kg) are dropped from heights of 16 m and 25 m.
respectively. The ratio of the time taken by them
to reach the ground is
Nor Alo​

Answer 1

Answer:

Given:

Object A is of mass 1 kg, Object B is of mass 3 kg. They are dropped from heights 16 m and 25 m respectively.

To find:

Ratio of time taken to reach the ground.

Concept:

Gravitational acceleration acts downwards on both the bodies. We shall assume that gravitational acceleration will remain constant all along the course of the bodies. So we can easily apply the equation of kinematics.

Calculation:

For 1st body , we can say :

$s1 = \dfrac{1}{2} g {(t1)}^{2}$

$= > 16 = \dfrac{1}{2} g {(t1)}^{2}$

For 2nd body , we can say :

$s2 = \dfrac{1}{2} g {(t2)}^{2}$

$= > 25 = \dfrac{1}{2} g {(t2)}^{2}$

Ratio of t1 and t2 :

$\therefore \: \dfrac{ {(t1)}^{2} }{ {(t2)}^{2} } = \dfrac{16}{25}$

$= > \: \dfrac{ (t1) }{ (t2)} = \dfrac{4}{5}$

So t1 : t2 = 4 : 5

Answer 2

Answer :

• Ratio is 4:5

Explanation :

For Body A

• Mass = 1 kg
• Height = 16 m
• Initial velocity (u) = 0
• Acceleration (a) = 10 m/s²

Use 2nd equation of motion :

$\longrightarrow \sf{s \: = \: ut_1 \: + \dfrac{1}{2} at_1 ^2} \\ \\ \longrightarrow \sf{16 \: = \: 0 \: + \: \dfrac{1}{2} \: \times \: 10 \: \times \: t_1 ^2} \\ \\ \longrightarrow \sf{16 \: = \: \dfrac{5t_1 ^2}{1}} \\ \\ \longrightarrow \sf{t_1 ^2 \: = \: \dfrac{16}{5}} \\ \\ \longrightarrow \sf{t_1 ^2 \: = \: 3.2} \\ \\ \longrightarrow \sf{t_1 \: = \: \sqrt{3.2} \: s}$

$\rule{100}{1}$

For Point B

• Mass = 5 kg
• Height = 25 m
• Initial velocity (u) = 0
• Acceleration (a) = 10 m/s²

$\longrightarrow \sf{s \: = \: ut_2 \: + \: \dfrac{1}{2} at_2 ^2} \\ \\ \longrightarrow \sf{25 \: = \: 0 \: + \: 5t_2 ^2} \\ \\ \longrightarrow \sf{t_2 ^2 \: = \: 5} \\ \\ \longrightarrow \sf{t_2 \: = \: \sqrt{5} \: s}$

______________________________

Now, ratio of their Time Period

$\longrightarrow \sf{\dfrac{t_1}{t_2} \: = \: \dfrac{\sqrt{3.5}}{\sqrt{5}}}$

Multiply by √5

$\\ \\ \longrightarrow \sf{\dfrac{t_1}{t_2} \: = \: \dfrac{\sqrt{3.4} \: \times \: \sqrt{5}}{\sqrt{5} \: \times \: \sqrt{5}}} \\ \\ \longrightarrow \sf{\dfrac{t_1}{t_2} \: = \: \dfrac{4}{5}} \\ \\ \underline{\underline {\sf{Ratio \: is \: 4 \: : \: 5}}}$