Question

A body is dropped from a height 122.5m.If it is stopped after 3 seconds and again released, further time of descent

Answer 1

Answer:

The time of descent is 4 sec.

Explanation:

Given that,

Distance d = 122.5 m

Time = 3 sec

We need to calculate the distance in 3 sec

In 3 sec the body descent

$s =ut+\dfrac{1}{2}gt^2$

Where, s = distance

t = time

g = acceleration due to the gravity

u = initial velocity

Put the value in the equation

$s=0+\dfrac{1}{2}\times9.8\times9$

$s=44.1\ m$

Now, The distance left

$s' = d-s$

$s' =122.5-44.1$

$s'= 78.4\ m$

Now, Time taken further of descent

Using equation of motion

$s'=ut+\dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s'}{g}}$

Where, s' = left distance

$t=\sqrt{\dfrac{2\times78.4}{9.8}}$

$t=4\ sec$

Hence, The time of descent is 4 sec.

Answer 2

Answer:

Hope this helps

Explanation: