Respuesta :
The ball launched up a semicircular chute is [tex]2R -\dfrac{1}{2}gt^2[/tex] far from the bottom of the chute.
Given to us
Centripetal acceleration = 2g,
y₀ = 2R
[tex]a_y[/tex] = -g
What is the Velocity of the ball?
We know that the centripetal acceleration can be written as,
[tex]\alpha =\dfrac{\rm v^2}{R}[/tex]
Substituting the value of α,
[tex]2g =\dfrac{\rm v^2}{R}\\\\\rm v^2 = R2g[/tex]
Time of free fall
We can write the formula for the time of free fall,
[tex]y(t) = y_0 +\rm V_0_yt +\dfrac{1}{2}a_yt^2\\\\[/tex]
substitute the values we get,
[tex]y(t) = 2R + 0 +\dfrac{1}{2}(-g)t^2\\\\y(t) = 2R -\dfrac{1}{2}gt^2[/tex]
Hence, the ball is [tex]2R -\dfrac{1}{2}gt^2[/tex] far from the bottom of the chute.
Learn more about Centripetal acceleration:
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