(Mechanics)An interesting question about projectile

[physicsisgrea]

Homework Statement

Suppose an elastic ball is set into projectile motion on an inclined plane, which makes an angle of 30 degrees to the horizontal. It is projected with an initial velocity of 5 m/s, making an angle of 20 degrees to the slope of the plane. Suppose the plane is very long, and no energy loss throughout.

a. Find the time just after the 10th bounce, assume time = 0 when the ball is projected. (g = 10 m/s^2).
b. Hence find the distance traveled by the ball.

Homework Equations

x = uxt
y = uyt - .5gt^2
and other equations about projectile motion

The Attempt at a Solution

t' = 2(5 cos 20) / (10 sin 30) = 1.8794 s
So T = 10(1.8794) = 18.8 s.

However the answer is 3.95 s, far smaller than mine.

My instructor gives the answer as follows:
t' = 2(5sin20) / (10cos30) = 0 = 0.395 s
So T = t' * 10 = 3.95 s
I don't really understand what he is doing, and can anyone explain the approach of my instructor?

 

[tiny-tim]

hi physicsisgrea!

show us how you got from …

x = uxt
y = uyt - .5gt^2

to …

t' = 2(5 cos 20) / (10 sin 30) = 1.8794 s

… and we'll se what went wrong

 

[physicsisgrea]

hi physicsisgrea!

show us how you got from …to …… and we'll se what went wrong

oh sorry!
y = u sin 20 t - .5 gt^2 cos 30
x = u cos 20 t + .5 gt^2 sin 30

are they correct?

 

[tiny-tim]

looks ok…

and then?​