Aball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 g. how far from the bottom of the chute does the ball land?

The ball launched up a semicircular chute is far from the bottom of the chute.

Given to us

Centripetal acceleration = 2g,

y₀ = 2R

= -g

We know that the centripetal acceleration can be written as,

Substituting the value of α,

We can write the formula for the time of free fall,

substitute the values we get,

Hence, the ball is far from the bottom of the chute.

Learn more about Centripetal acceleration:

link

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.