A particle P falling vertically from a height hits a plane inclined to the horizontal at an angle θ with speed v and rebounds elastically, as shown. The distance along the plane where it hits the second time is

Question

A particle P falling vertically from a height hits a plane inclined to the horizontal at an angle θ with speed v and rebounds elastically, as shown. The distance along the plane where it hits the second time is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/038dc73541323c00acdef1e47538ce19-.png" /> (A)  (4v2 sin2 θ/g)  (B)  (v2 sin2 θ/2g)  (C)  (4v2 sin θ/g)  (D)  (v2 sin θ/4g)

Tardigrade Admin · Accepted Answer

We have two components for each initial velocity

(v_{0})

and acceleration due to gravity

(g)

. In this frame of reference the horizontal and vertical component of velocity and gravity are given by

V_{x} = V_{0} sin θ, v_{y} = V_{0} cos θ,

g_{x} = g sin θ, g_{y} = g cos θ

Now,

t_{b o u n ce} = \frac{2 V _{y}}{g _{y}} = \frac{2 V _{0} c o s θ}{g c o s θ} = \frac{2 V _{0}}{g}

For horizontal distance

d = v_{x} t + \frac{1}{2} g_{x} t^{2}

∴ d = v_{0} sin θ (\frac{2 V _{0}}{g}) + \frac{1}{2} g sin θ [\frac{2 V _{0}}{g}]^{2}

\frac{= 2 v _{0}^{2} s i n θ}{g} + \frac{1}{2} g \frac{4 V ^{2}}{g ^{2}} sin θ = \frac{4 V _{0}^{2} s i n θ}{g}

Q. A particle $P$ falling vertically from a height hits a plane inclined to the horizontal at an angle $θ$ with speed $v$ and rebounds elastically, as shown. The distance along the plane where it hits the second time is

$\frac{4 v ^{2} s i n ^{2} θ}{g}$

$\frac{v ^{2} s i n ^{2} θ}{2 g}$

$\frac{4 v ^{2} s i n θ}{g}$

$\frac{v ^{2} s i n θ}{4 g}$

Q. A particle P falling vertically from a height hits a plane inclined to the horizontal at an angle θ with speed v and rebounds elastically, as shown. The distance along the plane where it hits the second time is

g4v2sin2θ​

2gv2sin2θ​

g4v2sinθ​

4gv2sinθ​

Q. A particle $P$ falling vertically from a height hits a plane inclined to the horizontal at an angle $θ$ with speed $v$ and rebounds elastically, as shown. The distance along the plane where it hits the second time is

$\frac{4 v ^{2} s i n ^{2} θ}{g}$

$\frac{v ^{2} s i n ^{2} θ}{2 g}$

$\frac{4 v ^{2} s i n θ}{g}$

$\frac{v ^{2} s i n θ}{4 g}$