Question

A body is released from the top of the tower of height 100m . After 2 seconds it is stopped and then instantaneously released. what will be its height after next 3 seconds? [Take g=10m/s ^-2]
1》40m
2》35m
3》45m
4》30m​

Answer 1

Given:

A body is released from the top of the tower of height 100m . After 2 seconds it is stopped and then instantaneously released.

To find:

Height of body after next 3 seconds.

Calculation:

After 2 seconds , the height of the ball above the earth will be:

$\therefore \: h = 100 - y$

$= > \: h = 100 - \bigg \{ut + \dfrac{1}{2} g {t}^{2} \bigg \}$

$= > \: h = 100 - \bigg \{0 + \dfrac{1}{2} g {t}^{2} \bigg \}$

$= > \: h = 100 - \bigg \{ \dfrac{1}{2} g {t}^{2} \bigg \}$

$= > \: h = 100 - \bigg \{ \dfrac{1}{2} g {(2)}^{2} \bigg \}$

$= > h = 100 - 20$

$= > h = 80 \: m$

Its stopped for an instant and again released.

Let height after next 3 seconds be H.

$\therefore \: H = 80 -( y2)$

$= > \: H = 80 - \bigg \{u(t2) + \dfrac{1}{2} g {(t2)}^{2} \bigg \}$

$= > \: H = 80 - \bigg \{0 + \dfrac{1}{2} g {(t2)}^{2} \bigg \}$

$= > \: H = 80 - \bigg \{ \dfrac{1}{2} g {(t2)}^{2} \bigg \}$

$= > \: H = 80 - \bigg \{ \dfrac{1}{2} g {(3)}^{2} \bigg \}$

$= > \: H = 80 - \bigg \{ 45\bigg \}$

$= > \: H = 35 \: m$

So, final answer is :

$\boxed{ \red{ \bold{ \large{ \: H = 35 \: m}}}}$