Let the angle be θ
Assume that h be the height above the point of impact that the ball is dropped.
From Newton's third equation of motion
v² = u² + 2as
v² = 0 + 2(-9.8)(-h)
v = √(19.6h)
Also since we are dealing in a perfectly elastic collision
angle of incidence = angle of reflection
angle of incidence = (90-θ) degrees above slope
therefore angle of reflection = (90-θ) degrees above slope.
The speed of the ball will be the same as before, as it is perfectly elastic.
x-component of velocity = √(19.6h) cos(90-θ)
y-component of velocity = √(19.6h) sin(90-θ)
taking the point of impact to be the origin and the slope now to be horizontal.
x = (√(19.6h) cos(90-θ))t
= (√(19.6h) sin(θ))t
y = (√(19.6h) sin(90-θ))t + 0.5(-9.8)t²
= (√(19.6h) cos(θ))t + 0.5(-9.8)t²
his slope when y = 0
0 = (√(19.6h) cos(θ))t + 0.5(-9.8)t²
t = 0 (initially when ball hits the slope)
t = (√(19.6h) cos(θ)) / 4.9
subbing into x:
x = ((√(19.6h) cos(θ)) / 4.9)((√(19.6h) sin(θ)))
x = (19.6h /4.9) cos(θ) sin(θ)
putting this into matrix form and doing a matrix rotation of angle θ clockwise.
x' = x cos(-θ) - y sin(-θ) = x cos(θ) + y sin(θ)
y' = x sin(-θ) + cos(-θ) = -x sin(θ) + y sin(θ)
x' = (19.6h /4.9) cos²(θ) sin(θ)
y' = -(19.6h /4.9) cos²(θ) sin²(θ)
these are the coordinates for where on the slope it will hit again, taking the origin at the point of impact.
Let the velocity with which the ball hits the
Plane be v .
Before the collision v perpendicular to plane is -v cos(@).
After the collision v perpendicular to plane is +v cos(@).
Before the collision velocity parallel to plane is -v sin(@) which remains unchanged even after the collision.
For the ball to hit the plane displacement n the direction perpendicular to the plane should be zero while the acceleration perpendicular to plane is -g cos(@).
By writing the kinematics equation n direction perpendicular to plane... After time t;
0= v cos(@)*t- 1/2 (g cos@)*t²
=> t= 2*(v/g) which is the time after which the ball hits the plane.
Now displacement n direction parallel to plane is -l(say) velocity parallel to plane is
-v sin(@) and acceleration in this direction is
- g sin(@)
Writing the equation of displacement parallel to plane...
l= v sin(@)*t + 1/2(g sin@)*t²
Where t= 2*(v/g)
=> l= 4*(v²sin(@)/g) this is the displacement of ball parallel to the plane which makes angle @ with horizontal.
=> Component of displacement along horizontal is l*cos(@)=x(say)
Component of displacement perpendicular to horizontal i.e along vertical direction is l*sin(@)=y(say)
Thank you :)