A particle is projected horizontally with speed u from point A which is 10 m above the ground. If the particle hits the inclined plane perpendicularly at point B.
(a) Find the horizontal speed with which the particle was projected
(b) Find the length OB along the inclined plane
(c) Speed at point B will be
Answers
Answer:
Taking positive x axis parallel to OB and y axis perpendicular to it (downward as positive)
Initial condition
under initial condition
ux = ucos 450
uy = usin 45o
ax = -gsin45o
ay = gcos45o
to determine t
using equation of motion along x axis
vx = ux +axt
as vx = 0
so t = -(ux/ax)
or, t = ucos 45o/gsin45o
t = u/g
this is the time of flight
now using it
we drop a perpendicular from projection point on OB
the length of the projection is l = 10 sin45o
using equation of motion along y axis
y = uy + ayt2/2
10sin 45o = usin45o + 12gcos 45o(u/g)2
we get 10 = u + u2/2g
or u2 + 2ug = 20g
or, u2 +20u - 200 = 0
solving this quadratic equation we get,
u = 7.32 m/s
at point B , it has only velocity in y direction so its velocity in x direction has become zero
hence using the equation of motion along y axis
vx2 = ux2 + 2ax
as vx = 0
so x = -(u cos45o)2 / 2(-gsin 45o)
x = u2/22g
x=7.32220×2x=53.5828.28x=1.89 m
so using equation of motion along y axis
vy = uy +ayt
vy = usin 45o + u(gcos450)/g
vy = usin45o + ucos450
thus total velocity at B is vy as vx = 0
v=7.3212+12v=7.32×2=10.35 m/