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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
Option: 1
$2h \cos (\alpha /2)$

Option: 2
$h/2 \sin^2 \alpha$

Option: 3
$2h \sin (2\alpha)$

Option: 4
$2h \cot (\alpha /2)$

Answers (1)

best_answer

As we have learned

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

$R=\frac{u^{2}\sin 2\Theta }{g}$

- wherein

Special case of horizontal range

For man horizontal range.

$\Theta = 45^{0}$

$R_{max}=\frac{u^{2}\sin 2 (45) }{g}$ $=\frac{u^{2}\times 1}{g}$ $=\frac{u^{2}}{g}$

Pm and Qn are two waves , just clear by projectile

$y = x \tan \alpha - \frac{gx^2 }{2u^2\cos ^2 \alpha } .........(1)$

y coordinate of points P and Q must satisfy equation (1) i.e

$h= x \tan \alpha - \frac{gx^2}{2u^2 \cos ^{2}\alpha } \; \; or \; \; gx^2-2u^2x\sin \alpha \cos \alpha + 2h u^2\cos ^2 \alpha = 0$

Let x1 and x2 be coordinates of points p and q the x1 and x2 are the roots of quadratic equation (2)

$\Rightarrow x_1 + x_2= \frac{2u^2 \sin \alpha \cos \alpha }{g}.....(3)$

and

$\Rightarrow x_1 x_2= \frac{2hu^2 \cos^2 \alpha }{g}.....(4)$

Let R be the range of porjectile

i.e OB= R .From the symmetry of the path about axis of parabola , we have

OB = ON+NB= x1 +x2

$R = \frac{2u^2 \sin \alpha \cos \alpha }{g}.....(5)$

Distance between the wave = $2h = x_2-x_1$

$(x_2-x_1)^2= (x_2+x_1)^2- 4x_1x_2$

$\Rightarrow 4h^2= R^2 - \frac{8hu^2\cos ^{2}\alpha }{g} ....(6)$

$R^2 -4hR \cot \alpha -4h^2= 0 (from \; \; 5\; \; and \; \; 6 )$

$\Rightarrow R = \frac{4h \cot \alpha \pm \sqrt{16h^2 \cot^2 \alpha+16 h^2}}{2}$

$\Rightarrow R = \frac{4h \cot \alpha +4h \csc \alpha }{2}=$

$2h (\cot \alpha + \csc \alpha )= 2h\left ( \frac{1+\cos \alpha }{\sin \alpha } \right )$

$R= 2h \cot \alpha /2$

Posted by

HARSH KANKARIA

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