Question

Question:-
(Refer to the attachment.)
________...
IXth Physics.

Answer 1

Given thαt,

• Height(h) = $\displaystyle{\sf{ \dfrac{1}{2}gT^{2}}}$ or $\displaystyle{\sf{\dfrac{h}{2}}}$

❍ We need to find the time tαken in sec.

_________

Let's assume thαt time be T.

According to the question,

$\displaystyle{\sf{h = \dfrac{1}{2}gT^{2}}}$

• Therefore, h/2 is αctuαlly -

$\displaystyle{\sf{ \dfrac{h}{2} = \dfrac{1}{2}g(T-1)^{2}}}$

• (T-1)² (becαuse the lαst pαrt wαs for distαnce and this is for time.)
• Substitute the vαlues and simplify.

$\implies{\sf{ \dfrac{\cancel{\frac{1g}{2}}}{2}T^{2}= \cancel{\dfrac{1}{2}g}(T-1)^{2}}}$

$\implies{\sf{\dfrac{T^{2}}{2}=(T-1)^{2}}}$

$\implies{\sf{T^{2} = 2(T-1)^{2}}}$

$\implies{\sf{ \dfrac{T^{2}}{(T-1)^{2}}=2}}$

$\implies{\sf{ \dfrac{T}{T-1} = \sqrt{2}}}$

$\implies{\sf{ T = \sqrt{2}T - \sqrt{2}}}$

$\implies{\sf{T = ( \sqrt{2} - 1)T = \sqrt{2}}}$

$\displaystyle{\sf{ T = \dfrac{ \sqrt{2}}{ \sqrt{2} - 1}}}$

• Rαtionαlize the denominαtor.

$\implies{\sf{ \dfrac{ \sqrt{2}}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}}}$

$\implies{\sf{ \dfrac{ \sqrt{2}(\sqrt{2}+1)}{(\sqrt{2} - 1)}}}$

$\large\implies{\boxed{\sf{\red{2+ \sqrt{2}}}}}$

Hence, the time of trαvel is 2 + √2 sec(option B).

And we αre done! :D

__________________________

Answer 2

Question

A body starts falling from height 'h' and travels distance h/2 during the last second of motion. The time of travel (in sec.) is-

Answer

Given :

A body starts falling from height 'h' and travels distance h/2 during the last second of motion

To Find :

The time of travel

Solution :

Here we have to use 2nd equation of motion

when the body travel h distance,

$\leadsto \sf h=ut+\dfrac{1}{2}gt^2$

$\leadsto \sf h=(0)t + \dfrac{1}{2}gt^2$

$\leadsto \sf h = \dfrac{1}{2}gt^2 \: \: \: \: \: ...(1)$

when the body travels h/2 distance

$\leadsto\sf \dfrac{h}{2}=ut+\dfrac{1}{2}g(t-1)^2$

$\leadsto\sf \dfrac{h}{2}=(0)t + \dfrac{1}{2}gt^2$

$\leadsto\sf \dfrac{h}{2}= \dfrac{1}{2}gt^2 \: \: \: \: \: .......(2)$

On dividing eq (1) and (2)

$\leadsto\sf \dfrac{h}{\dfrac{h}{2}}=\dfrac{\dfrac{1}{2}gt^2}{\dfrac{1}{2}g(t-1)^2}$

$\leadsto \sf 2=\dfrac{t^2}{(t-1)^2}$

$\leadsto \sf \sqrt{2} = \dfrac{t}{t-1}$

$\leadsto \sf t=\sqrt{2}(t-1)$

$\leadsto \sf t=\sqrt{2}t-\sqrt{2}$

$\leadsto\sf t(\sqrt{2}-1)=\sqrt{2}$

$\leadsto \sf t=\dfrac{\sqrt{2}}{\sqrt{2}-1}$

Now we have to Rationalize the denominator,

$\leadsto \sf t=\dfrac{\sqrt{2}}{{\sqrt{2}-1}}\times \dfrac{{\sqrt{2}+1}}{{\sqrt{2}+1}}$

$\leadsto \sf t=\dfrac{\sqrt{2}({\sqrt{2}+1})}{2-1}$

$\leadsto \underline{\boxed{ \purple{ \sf t=2+\sqrt{2}}}}$

Thus, Option (b) is correct answer.