Question

55. A body is dropped from a height H. The time taken to
cover second half of the journey is​

Answer 1

Answer:

• The time taken (t') to cover second half of distance is √ H / g (√2 - 1).

Given:

1. Total Height = H.
2. Let the total time taken to reach the ground is " T "

Explanation:

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The total time taken by the body to reach the ground i.e. to cover full Height.

From second kinematic equation,

⇒ S = u t + 1 / 2 a t²

Where,

• S denotes distance.
• u denotes initial velocity.
• t denotes time taken.
• a denotes acceleration.

Now,

⇒ H = u t + 1 / 2 g T²

⇒ H = 0 + 1 / 2 g T²   ∵ [u = 0 m/s]

⇒ H = 1 / 2 g T²

⇒ T = √ 2 H / g __[1]

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The total time taken by the body to reach the half of the height i.e. to cover H / 2.

From second kinematic equation,

⇒ S = u t + 1 / 2 a t²

Where,

• S denotes distance.
• u denotes initial velocity.
• t denotes time taken.
• a denotes acceleration.

Now,

⇒ 1 / 2 H = u t + 1 / 2 g t²

⇒ 1 / 2 H = 0 + 1 / 2 g t²   ∵ [u = 0 m/s]

⇒ 1 / 2 H = 1 / 2 g t²

⇒ T = √  H / g __[2]

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Now, to get time taken to cover second half of distance is given by,

⇒ t' = T - t

Where,

• T is total time to cover Height " H "
• t is time to cover Height " H / 2 "
• t' is time to cover Second half of journey.

Now,

⇒ t' = T - t

⇒ t' = √ 2 H / g - √ H / g

⇒ t' = √  H / g (√2 - 1)

⇒ t' = √  H / g (√2 - 1)

∴ The time taken (t') to cover second half of distance is √ H / g (√2 - 1).

Hence, Option - 3 is correct.

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