Question

a freely falling body takes 't' seconds to travel first (1/x ) th distance. then its time descent is. answer plz.. step by step​

Answer 1

Answer:

Explanation:

u=0,a=g

Let the total height be H

Time of descent = T

If H is the total height then 1/x of total height is H/x

By second equation of motion

S=ut+1/2at²

1/x=1/2gt²

H/x=1/2gt²

H=gt²x/2 .........(1)

Now taking distance as H, the time will the total time of descent which is T.

Applying second equation of motion

H=ut+1/2gT²

H=0+1/2gT²

From(1),

gt²x/2=gT²/2

T²=xt²

T=t√x

Hence the time taken to descend is t√x

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Answer 2

Answer:

• The time of decent is (t_d) is t √ x seconds.

Given:

1. Time taken to travel 1 / x distance is "t"
2. Lets take total distance as total height i.e. 'H'
3. Let the time of descent be denoted by t_d
4. Initial velocity = 0 m/s.

Explanation:

$\rule{300}{1.5}$

From the given statement.

Freely falling body takes 't' seconds to travel first (1/x ) th distance.

As we assumed total distance as height (H)

Therefore,

⇒ 1 / x of Distance = 1 / x × H

⇒ 1 / x of Distance = H / x

⇒ S = H / x

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying the second kinematic equation,

(For H / x Distance which it covers in 't' seconds)

⇒ S = u t + 1 / 2 a t²

Where,

• S denotes Distance.
• u Denotes Initial velocity.
• t Denotes time.
• a Denotes acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ H / x = 0 × t + 1 / 2 × g × (t)²

∵ [ S = H / x ; t = t ; u = 0 ; a = g ]

⇒ H / x = 0 + 1 / 2 × g × t²

⇒ H / x = 1 / 2 × g × t²

⇒ H / x = 1 / 2 g t²

⇒ H = x × (1 / 2 g t²)

⇒ H = x × g t² / 2

⇒ H = ( x g t² ) / 2

∴ We get the Height through which the body falls.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now,

Applying the second kinematic equation,

(For H  Distance which it covers in 't_d' seconds)

Note:- H denotes total height in which it covers time of decent.

⇒  S = u t + 1 / 2 a t²

Where,

• S denotes Distance.
• u Denotes Initial velocity.
• t Denotes time.
• a Denotes acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ H = 0 × t_d + 1 / 2 × g × (t_d)²

∵ [ S = H  ; t = t_d ; u = 0 ; a = g ]

⇒ H = 0 +  1 / 2 × g × t_d²

⇒ x g t² / 2 = g t_d² / 2

∵ [ H = x g t² / 2 ]

⇒ x g t² = g t_d²

⇒ x t² = t_d²

⇒ t_d² = x t²

⇒ t_d = √ x t²

⇒ t_d = t √ x

⇒ t_d = t √ x seconds.

∴ The time of decent is (t_d) is t √ x seconds .

$\rule{300}{1.5}$