Question

solve this problem its important​

Answer 1

Answer:

√gh...

dropped body--u=0

s=1/2gt^2

h/2=1/2gt^2

t=√h/g    ⇒ 1

vertically projected---

s=ut-1/2gt^2

h/2=v√h/g-[g/2(h/g)]         (∵from 1)

⇒v=h√g/h

  =√gh

 

Explanation:

Answer 2

Answer:

Given:

Body A thrown vertically upwards with velocity

v_{0} and body B dropped from height H. They meet at height H/2

To find:

Value of v_(0)/2

Calculation:

For the body B , we can say that :

$\boxed{ \large{ \bold{ \dfrac{H}{2} = \dfrac{1}{2} g {t}^{2} }}}$

For body A , we can say that :

$\boxed{\large{ \bold{ \dfrac{H}{2} = v_{0}t - \frac{1}{2}g {t}^{2} }}}$

Adding the 2 Equations , we get :

$\large{ \bold{ \therefore \dfrac{H}{2} + \dfrac{H}{2} = v_{0}t }}$

$\large{ \bold{ = > H = v_{0}t }}$

$\bold{ \large{= > v_{0}= \dfrac{H}{t} }}$

Putting the value of t = √(h/g) :

$\bold{ \large{= > v_{0}= \dfrac{H}{ \sqrt{ \frac{H}{g} } } }}$

$\bold{ \large{= > v_{0}= \sqrt{gH} }}$

So final answer :

$\boxed{\red{ \bold{ \huge{ v_{0}= \sqrt{gH} }}}}$