Question

can any one solve this ??

Answer 1

Given,

Height ( s ) = h

Time ( t ) = t1

Initial velocity ( u ) = 0


Using Second equation of Motion,

⇒ s = ut + ( 1/2 )at²

⇒ h = 0 × t1 + 0.5a( t1 )²

⇒ h = 0 + 0.5 a t1²

⇒ h = 0.5 a t1²

⇒ h/ ( 0.5 t1² ) = a

⇒ h/ [ ( 1/2 ) t1² ] = a

⇒ 2h/t1² = a -------- ( 1 )

Now, we have to find the time taken to cover the first half of the distance.

So,

Initial velocity ( u ) = 0

Distance ( s ) = h/2

Acceleration ( a ) = ( 2h/t1² )

Time ( t2 ) = ?

Using second equation of Motion,

⇒s = ut + ( 1/2 ) at²

⇒ ( h/2 ) = 0 × t2 + ( 1/2 ) ( 2h/t1² ) ( t2 )²

⇒ ( h/2 ) = 0 + ( h/t1² ) t2²

⇒ ( h/2 ) = ( h/t1² )t2²

⇒ [ h / ( 2 × h ) ] = ( 1/t1² )t2²

⇒ ( 1/2 ) = ( t2² ) / ( t1² )

Using identity,

⇒ [ ( a^m / b^m ) ] = ( a/b )^m

⇒ ( 1/2 ) = ( t2 / t1 )²

⇒ √( 1/2 ) = t2/t1

⇒ 1/√2 = t2/t1


⇒ t1 /√2 = t2

•°• t2 = t1/√2

Hence, the required answer is option ( 1 ).

Answer 2

Hey friend, Harish here.

Here is your answer.

$\mathrm{We\ know\ that\ :} \\ \\ \to \mathrm{h=ut+\frac{1}{2}gt^2} \\ \\ \mathrm{Here, u= 0\ m/s \ because \ it\ is\ just\ dropped.}\\ \\ \mathrm{Then,} \\ \\ \to \mathrm{h= \frac{1}{2}g(t_1)^2} \\ \\ \to \mathrm{(t_1)= \sqrt{ \frac{2h}{g} }} \ \ -(i) \\ \\ \to \mathrm{\frac{h}{2} = \frac{1}{2}g(t_2)^2} \\ \\ \to \mathrm{(t_2)=\sqrt{\frac{2h}{2g}}= \sqrt{\frac{h}{g}}} \ \ - (ii) \\ \\ \mathrm{(i)\ Divided\ by\ (ii)\ we\ get\ :} \\ \\ \to \mathrm{\frac{t_1}{t_2} = \sqrt{\frac{\frac{2h}{g}}{\frac{h}{g}}}}$

$\to \mathrm{\frac{t_1}{t_2}=\frac{\sqrt{2}}{1}} \\ \\ \to \boxed{\bold{\mathrm{t_2 = \frac{t_1}{\sqrt{2}}}}}$
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Hope my answer is helpful to you.