Question

A body is projected at an angle alpha to the horizontal so as to clear two walls of equal height 'a' at a distance '2a' from each other .show that the range equal to 2acot(apha/2)​

Answer 1

Answer:

$R_{min} = 2a cot(\frac{\alpha}{2})$

Explanation:

Correction : To prove : Minimum Range = $2a cot(\frac{\alpha}{2})$

Let angle made by projectile on the wall from horizontal be u2 and angle from horizontal be $\theta$ as shown in figure.  

Now , we know thatHorizontal component :

$u_1 \cos(\alpha) = u_2 cos(\theta)$ .

Range from wall to wall :

$2a = \frac{u_2^2sin(2\theta}{g}$

Height :

$a+\frac{u_2^2 sin^2(\theta)}{2g} =\frac{u_1^2 sin^2(\alpha)}{2g}$

Using above equations :  

we get ,  

$u_1^2 - u_2^2 =u_2^2 sin(2\theta)$

$=> \frac{u_1^2}{u_2^2}-1 = sin(2\theta)$

$=>\frac{cos^2(\theta)}{cos^2(\alpha)}-1 = sin(2\theta)$

$=> cos(\alpha) = \frac{\cos(\theta)}{sin(\theta)+cos(\theta)}$  

$=> tan(\theta) = sec(\alpha) - 1$  

Now ,

$Range (min) =\frac{u_1^2sin(2\alpha)}{g}$

Using equation of horizontal component and range from wall to wall to get :  

$Range (min) = \frac{2a tan(\alpha)}{tan(\theta)}$

  $= 2a \frac{tan(\alpha)}{sec(\alpha) - 1 )}$  

                     $= 2a cot(\frac{\alpha}{2})$

Answer 2

Answer:

yrrr kaha ho

Ghar jaa KR bhul hi gye