A ball is launched up a semicircular tube of radius R (see Fig.1) in such a way

Question

A ball is launched up a semicircular tube of radius R (see Fig.1) in such a way that at the top of the tube, just before it goes into free fall, the ball has a centripetal acceleration of 2g.

a. Determine the magnitude and direction of the initial velocity v0 of the ball at the beginning of the free fall.

b. Determine the equation y(x) describing the trajectory of the ball while in the air.

c. Determine the range L of the ball.

d. Determine the time of flight T of the ball.

e. Determine the final velocity v f (as a vector) when the ball hits the ground.

Explanation / Answer

In The centripetal acceleration is:
a_c = v^2/R , so:
v^2/R = a_c = 2*g
v = sqrt(2*g*R)

v is solve it for put that value

Once it escapes the chute, it's got the following initial conditions:
x(0) = 0
vx(0) = sqrt(2gR)
=> x(t) = x(0) + vx(0)*t = sqrt(2gR)*t

y(0) = 2R
vy(0) = 0
=>
y(t) = y(0) + vy(0)*t -gt^2/2 = 2R - g*t^2/2

It hits the ground at t = T , when y(T) = 0:
0 = 2R - g*T^2/2
=> T = sqrt(4R/g)

At that point,
x(T) = sqrt(2gR)*sqrt(4R/g) = sqrt(2gR*4R/g) = 2R*sqrt(2)

So it lands at x = 2*sqrt(2)*R

that's a land at x.

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