Question

A body of mass 'm' is dropped from height '1m' and body B of mass '3m' is dropped from height '9m'. calculate the ratio of time taken by the bodies A and B to reach the ground (g=10ms)​

Answer 1

Answer:

In first case, A body , s= 1 m, mass= m, initial velocity =u=0,

$s = ut + \frac{1}{2} a {t}^{2} \\ 1= u \times t + \frac{1}{2} g {t}^{2} ..........(as \: acceleration \: is \: due \: to \: gravity) \\ 1= 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ 1= \frac{1}{2} \times 10 \times {t}^{2} \\ 1 = 5 \times {t}^{2} \\ {t}^{2} = \frac{1}{5} \\ t = \sqrt{ \frac{1}{5} } \\ t = \frac{1}{ \sqrt{5} } m$

In second case,B object , s= 9m ,mass=3m initial velocity= u = 0

$s = ut + \frac{1}{2} g {t}^{2} ......(acceleration \: due \: to \: gravity) \\ 9 = 0 \times t \frac{1}{2} \times 10 {t}^{2} \\ 9 = \frac{1}{2} \times 10 {t}^{2} \\ 9 = 5 {t}^{2} \\ {t}^{2} = \frac{9}{5} \\ t = \sqrt{ \frac{9}{5} } \\ t = \frac{ \sqrt{9} }{ \sqrt{5} } m$

Now ratio of both will be:

$\frac{1}{ \sqrt{5} } is \: to \: \frac{ \sqrt{9} }{ \sqrt{5} }m$

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