Question

2. A body is dropped from a height H. The time taken to
cover second half of the journey is
2H
H
(1) 29
(2) va​

Answer 1

Answer:

• The required time taken to cover second half of the journey is √H/g(√2 - 1).

Explanation:

We have been given that a body is dropped from a height H. It means we can use “free fall of body condition” concept.

Here, We have some points for the first condition;

• Initial Velocity(u) = 0
• Acceleration (g) = g
• Time = t

We will second motion of equation which is given by,

• S = ut+½at²

Here, S will be replaced by H. Now, We have a new relation.

H = (0)t + ½gt²

H = ½gt²

∵ t = √(2H/g) .....(i)

Case: 2

For the second case, We have the following data.

• Intial Velocity (u) = 0
• Acceleration (g) = g
• Time = T (By assumption)
• Height covered by body = H/2 (By given condition)

S = uT+½aT²

H/2 = (0) + ½T²

∵ T = √H/g .....(ii)

Now, According to the Question's Statement:

• The time taken to
• cover second half of the journey given by ( t - T ) is equal to T' .

So,We have an expression.

T' = √2H/g - √H/g

∵ T' = √H/g(√2 - 1)

• Hence, the required time taken to

cover second half of the journey is √H/g(√2 - 1).

Answer 2

Answer:

Distance covered by body under free fall condition,

where,

=> Initial velocity = 0

=> acceleration = g

By using 2nd equation in terms of gravitation,

$\sf{\implies H = ut+\dfrac{1}{2} gt^{2}}$

$\sf{\implies H = 0\times t+\dfrac{1}{2} (g)t^{2}}$

$\sf{\implies H = \dfrac{1}{2}gt^{2}}$

$\sf{\implies t=\sqrt{\dfrac{2H}{g}}\;\;\;\;(time\;to\;cover\;height\;H)}$

Let at t1 body covers first H/2

$\sf{t_{1}=\sqrt{\dfrac{H}{g}}}$

So,

$\sf{\implies t-t_{1}=\sqrt{\dfrac{2H}{g}} -\sqrt{\dfrac{H}{g}}}$

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

${\boxed{\boxed{\sf{So,\;time\;taken\;to\;cover\;second\;half=\sqrt{\dfrac{H}{g}}(\sqrt{2}-1)}}}}$