Question

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

Answer 1

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/