Question

Please answer this question ​

Answer 1

Answer:

$\sf u=\sqrt{\dfrac{5}{2} \ gh}$

Explanation:

We have equation of trajectory :

$\sf y=x. \ \tan \theta-\dfrac{g \ x^2}{2 \ u^2 \ \cos^2 \theta}$

Given a common point i.e. ( h , h ) :

$\sf h=h. \ \tan \theta-\dfrac{g \ h^2}{2 \ u^2 \ \cos^2 \theta}$

$\sf 1=\tan \theta-\dfrac{g \ h}{2 \ u^2 \ \cos^2 \theta}$

$\sf u^2.\cos^2 \theta=\dfrac{g \ h}{2(\tan \theta -1)}$

$\sf u=\dfrac{1}{\cos \theta} .\sqrt{\dfrac{g \ h}{2(\tan \theta -1)}}$

We know :

$\sf H_{max} = R/2$

Also :

H / R = tan Ф / 4

Putting H = R / 2 we get :

= > H / 2 H = tan Ф / 4

= > tan Ф = 2

We know :

tan = P / B

Finding cos Ф by applying Pythagoras' theorem :

cos Ф = 1 / √ 5

Putting value of cos Ф and tan Ф to find velocity :

$\sf u=\dfrac{1}{1\sqrt{5}} .\sqrt{\dfrac{g \ h}{2(2 -1)}}$

$\sf u=\sqrt{5}.\sqrt{\dfrac{g \ h}{2(1)}}$

$\sf u=\sqrt{\dfrac{5}{2} \ gh}$

Therefore we get required velocity .