(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?​

 

[asteriatic]

I would first commend you for attempting to solve the problem and for using the correct equations for projectile motion. However, I would suggest that you review your understanding of these equations and how they apply to this specific scenario.

In this case, the ball is not being projected horizontally, but rather at an angle of 20 degrees to the slope of the plane. This means that the initial velocity in the x-direction is not 5 m/s, but rather 5*cos(20) m/s. Similarly, the initial velocity in the y-direction is not 0 m/s, but rather 5*sin(20) m/s.

Using these correct initial velocities, the time for the ball to reach the ground after the first bounce can be calculated using the equation y = uyt - 0.5gt^2, where uyt is the initial velocity in the y-direction and g is the acceleration due to gravity. This gives a time of 0.395 s, which is the same as your instructor's answer.

For the distance traveled, you can use the equation x = uxt, where uxt is the initial velocity in the x-direction. This gives a distance of 5*cos(20) * 0.395 = 1.961 m for the first bounce. To find the total distance traveled after 10 bounces, you can use the formula S = uyt * T, where S is the total distance traveled, uyt is the initial velocity in the y-direction, and T is the total time for 10 bounces. This gives a distance of 5*sin(20) * 3.95 = 9.756 m, which is the same as your instructor's answer.

In summary, it is important to carefully consider the initial conditions and how they affect the equations used in projectile motion problems. I would suggest reviewing your understanding of these equations and practicing more problems to solidify your understanding.