A50g ball is released from rest 1.0 above the bottom of thetrack
shown. it rolls down a straight 30o segment, thenback up
a parabolic segment whose shape is given byy=(1/4)x2 ,
where x and y are in m. how high will theball go on the right
before reversing direction and rolling backdown?

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

We need to calculate the velocity

Using conservation of energy

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

Put the value into the formula

Put the value into the formula

We need to calculate the maximum height of the ball

Using again conservation of energy

Here, h = y highest point

Put the value into the formula

Put the value of y in the given equation

Hence, The maximum height of the ball is 2 m.

one "light year" is the distance that light travels, through vacuum, in one julian year.   that's 365.25 days, of 24 hours each.