Question

20. 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​

Answer 1

Solution: -

Answer:

$\tt{\implies \dfrac{t_{1}}{t_{2}} = \dfrac{4}{5}}$

Explanation:

Given:

$\tt{m_{1}= 1\;kg}$

$\tt{m_{2}= 3\;kg}$

$\tt{h_{1}= 16\;m}$

$\tt{h_{2}= 25\;m}$

To Find: The ratio of the time taken by them to  reach the ground.

​

Let time taken by the two bodies be $\tt{t_{1} \; and \; t_{2},}$ respectively.

Let assume that both bodies were initially at rest.

Using 2nd equation of motion:

$\tt{s= ut + \dfrac{1}{2} at^{2}}$

For a free falling body, a = g

$\tt{\therefore h_{1} = 0 + \dfrac{1}{2}gt_{1}^{2} \;\;\;\;.......(1)}$

$\tt{\therefore h_{2} = 0 + \dfrac{1}{2}gt_{2}^{2} \;\;\;\;.......(2)}$

Dividing equation (1) by equation (2)

$\tt{\implies \dfrac{h_{1}}{h_{2}} = \dfrac{t_{1}^{2}}{t_{2}^{2}}}$

$\tt{\implies \dfrac{16}{25} = \dfrac{t_{1}^{2}}{t_{2}^{2}}}$

${\boxed{\boxed{\tt{\implies \dfrac{t_{1}}{t_{2}} = \dfrac{4}{5}}}}}$

Answer 2

Refers to attachment!!!