Question

A ball is thrown downward on the 30 degree inclined plane so that when it rebounds perpendicular to the incline it has a velocity of V_A = 40 ft/s. Determine the distance R where it strikes the plane at B.

Answer 1

First, using the law of reflection, the incident angle is equal to the reflection angle. Since the ball is reflected at 90 degrees from the surface of the inclined plane, then the angle of incidence is zero, and the angle of reflection is also zero which results in a 90 degree bounce. I will also assume a perfectly elastic collision with the surface of the plane since no e factor was given. So the speed on impact is 40 ft/second at an angle of 60 degrees w//r/t the horizontal.

So , we can determine where point B is and the angle of launch

sin(30)*R=y of impact
cos(30)*R= x of impact

y(t)=vy0*t-.5*g*t^2
vy(t)=vy0-g*t
at impact
vy(t)=-40*sin(60)

-40*SIN(60)=vy0-g*t

(vy0+40*sin(60))/g=t
and

sin(30)*R=vy0*(vy0+40*sin(60))/g-
.5*(vy0+40*sin(60))^2/g

2*g*sin(30)*R=2*vy0*(vy0+40*sin(60))-
(vy0+40*sin(60))^2

x(t)=40*cos(60)*t

cos(30)*R=
40*cos(60)*(vy0+40*sin(60))/g

g*cos(30)*R=
40*cos(60)*(vy0+40*sin(60))

divide
2*tan(30)*40*cos(60)*
(vy0+40*sin(60))=
2*vy0*(vy0+40*sin(60))-
(vy0+40*sin(60))^2

solve for vy0
23.1*vy0+800=
2*vy0^2+69.3*vy0-(vy0^2+69.3*vy0+1200)

23.1*vy0+800=vy0^2-1200

0=vy0^2-23.1*vy0-2000

vy0=57.735 ft/s

-40*SIN(60)=57.735-g*t
t=2.887 seconds
R=66.67 ft
check
y(t)=vy0*t-.5*g*t^2
y=33.33333
R*sin(30)=y
R=66.67 ft
that checks

x(t)=40*cos(60)*t
x=57.735
cos(30)*R= x of impact
R=66.67 ft