A small ball moves along a track and slides over a…
A small ball moves along a track and slides over a cylindrical surface BC of radius $R = 2\ \mathrm{m}$. At C, where the radius makes $\alpha = 45^\circ$ with the vertical, the ball leaves the surface. Find how far (horizontally) it flies.
Explanation of Method
A ball leaving a cylindrical surface.
Step 1: Speed at the point of leaving
The ball leaves the convex surface where the normal force drops to zero; there the radial part of gravity alone provides the centripetal force: \[\frac{mv_C^2}{R} = mg\cos\alpha \;\Longrightarrow\; v_C = \sqrt{gR\cos 45^\circ} \approx 3.72\ \mathrm{m/s}.\]
So the leaving speed is fixed by the geometry alone; the initial speed is not needed.
Step 2: Flight (assumed geometry)
The figure is not available. Take BC as a quarter cylinder standing on the ground, with its axis at ground level. Then C is $R\cos45^\circ = 1.41\ \mathrm{m}$ above the ground, and the ball leaves along the tangent, 45° below the horizontal:
\[ v_x = v_y = 2.63\ \mathrm{m/s}, \]
\[ 1.41 = 2.63\,t + 4.905\,t^2, \]
\[ t \approx 0.332\ \mathrm{s}. \]
\[l = v_x t \approx 0.87\ \mathrm{m} \ \text{(horizontally from C)}\]
Result
\[\boxed{\,l = v_x t \approx 0.87\ \mathrm{m} \ \text{(horizontally from C)}\,}\]
Remark
About the original solution. Its answer of 9 m cannot be obtained from the data: it used the range formula $v^2\sin2\alpha/g$ for a projectile launched upward at 45° from level ground, whereas a ball leaving a convex surface moves downward along the tangent, and it needed an initial speed that is not given. The key physics is that the leaving point fixes the speed, $v_C^2 = gR\cos\alpha$. The final distance depends on the height of C above the landing surface, which should be taken from the original figure.