Question

A particle is projected in an oblique manner on a horizontal surface such that just crosses 2 walls of height a and b respectively which are located at distances of b and a from the point of projection respectively. Then find the range of this projectile

Answer 1

Answer:

For vertically upward motion of a projectile,

y=(usinα)t−

2

1

gt

2

where α is the angle of projection with the horizontal.

or h=(usinα)t−

2

1

gt

2

or gt

2

−(2usinα)t+2h=0.....(i)

∴t=

2g

2usinα±

(4u

2

sin

2

α)−8gh

If two roots of quadratic Eq. (i) are t

1

,t

2

, then

t

1

=

2g

2usinα+

4u

2

sin

2

α−8gh

.....(i)

and t

2

=

2g

2usinα−

(4u

2

sin

2

α)−8gh

....(ii)

If particle crosses the walls at times t

1

and t

2

respectively, then time of flight t is t=

t

1

t

2

or t

2

=t

1

t

2

.....(iii)

where total time of flight is given by t=

g

2usinα

....(iv)

Putting (i), (ii) and (iv) in equation (iii), we get

(

g

2usinα

)

2

=

4g

2

(2usinα)

2

−(4u

2

sin

2

α−8gh)

or

g

2

4u

2

sin

2

α

=

4g

2

8gh

or 2u

2

sin

2

α=gh

Given, u=

2gh

∴2(2gh)sin

2

α=gh

or sin

2

α=

4

1

or sinα=

2

1

∴α=30

∘

Answer 2

The range of the projectile is

$\boxed{\frac{a^2+ab+b^2}{a+b}}$

Explanation:

Let the initial velocity of the projectile is u and angle of projection be θ

Then

Horizontal displacement

$x=u\cos\theta t$  ................ (1)

Vertical displacement

$y=u\sin\theta t-\frac{1}{2}gt^2$

Putting the value of t from equation (1)

$y=u\sin\theta\times \frac{x}{u\cos\theta}-\frac{1}{2}g(\frac{x}{u\cos\theta})^2$

$\implies y=\tan\theta x-\frac{1}{2}g(\frac{x^2}{u^2\cos^2\theta})$

$\implies y=\tan\theta x-\frac{g}{2u^2\cos^2\theta}x^2$  

Taking $\tan\theta = X$

And $\frac{g}{2u^2\cos^2\theta}=Y$

We get the equation as

$y=Xx-Yx^2$   ................ (3)

The path of the projectile will be a parabola

According to the question, the coordinates of two points on its path will be (a, b) and (b, a)

Putting these values in equation (3)

$b=Xa-Ya^2$

$\implies \frac{b}{a}=X-Ya$   ...... (4)

And $a=Xb-Yb^2$

$\implies \frac{a}{b}=X-Yb$   ........ (5)

Subtracting eq (4) from (5)

$\frac{a}{b}-\frac{b}{a}=Y(a-b)$

$\implies \frac{a^2-b^2}{ab}=Y(a-b)$

$\implies Y=\frac{(a-b)(a+b)}{ab(a-b)}$

$\implies Y=\frac{a+b}{ab}$

$\implies \frac{g}{2u^2\cos^2\theta}=\frac{a+b}{ab}$

$\implies \frac{2u^2\cos^2\theta}{g}=\frac{ab}{a+b}$  ..... (6)

Putting the value of Y in eq (4)

$\frac{b}{a}=X-\frac{(a+b)a}{ab}$

$\implies X=\frac{b}{a}+\frac{a+b}{b}$

$\implies X=\frac{b^2+ab+a^2}{ab}$

$\implies \tan\theta=\frac{a^2+ab+b^2}{ab}$     ..... (7)

We know that range of a projectile is given by

$R=\frac{u^2\sin2\theta}{g}$

Therefore, multiplying eq (6) and (7)

$\frac{2u^2\cos^2\theta}{g}\times\tan\theta=\frac{ab}{a+b}\times \frac{a^2+ab+b^2}{ab}$

$\implies \frac{2u^2\cos^2\theta}{g}\times\frac{\sin\theta}{\cos\theta}=\frac{a^2+ab+b^2}{a+b}$

$\implies \frac{2u^2\cos\theta\sin\theta}{g}=\frac{a^2+ab+b^2}{a+b}$

$\implies \frac{u^2\sin2\theta}{g}=\frac{a^2+ab+b^2}{a+b}$

$R=\frac{a^2+ab+b^2}{a+b}$

Hope this answer is helpful.

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