Question

A body is projected at an angle
alpha to the horizontal so as to clear two waves of equal height h at a distance 2 h from each other, the horizontal range of projectile is
A. 2h cos(a/2)

B. h/2 sin² a

C 2h sin(2a)

D. 2h cot(a/2)


​

Answer 1

Answer:

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

min

=2acot(

2

α

)

Explanation:

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

2

α

)

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

Now , we know thatHorizontal component :

u_1 \cos(\alpha) = u_2 cos(\theta)u

1

cos(α)=u

2

cos(θ) .

Range from wall to wall :

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

g

u

2

2

sin(2θ

Height :

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

2g

u

2

2

sin

2

(θ)

=

2g

u

1

2

sin

2

(α)

Using above equations :

we get ,

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

1

2

−u

2

2

=u

2

2

sin(2θ)

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

u

2

2

u

1

2

−1=sin(2θ)

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

cos

2

(α)

cos

2

(θ)

−1=sin(2θ)

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

sin(θ)+cos(θ)

cos(θ)

=> tan(\theta) = sec(\alpha) - 1=>tan(θ)=sec(α)−1

Now ,

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

g

u

1

2

sin(2α)

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

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

tan(θ)

2atan(α)

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

sec(α)−1)

tan(α)

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

2

α

)