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Home » A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.
CBSE | Class 11 | Motion in a Plane | NCERT | Physics
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.

Answer:

When the x and y axes are set in the manner depicted in the diagram, the missile moves from point O to point A.

y = 0

uy = vo cos θ

ay = -g cos θ

t = T

Using the kinematics for the y-axis

y = uyt + 1/2 ayt2

T = 2vo cos θ/g cos θ

Using kinematics for the x-axis

x = uxt + 1/2 axt2

Solving for the distance,

L =\left[\frac{2 v _{0}}{ g }\right] v _{0} \sin \theta+\frac{1}{2} g \sin \theta\left[\frac{2 v _{0}}{ g }\right]^{2}
L =\frac{2 v _{0}^{2}}{ g } \sin \theta+\frac{1}{2} g \sin \theta \cdot \frac{4 v _{0}^{2}}{ g ^{2}}
=\frac{2 v _{0}^{2}}{ g }[\sin \theta+\sin \theta]=\frac{2 v _{0}^{2}}{ g } 2 \sin \theta
\Longrightarrow L =\frac{4 v _{0}^{2}}{ g } \sin \theta

i

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