Question

part d now, switch over to the large ball drop tracker and observe the graphs of vertical displacement against time (y vs. t) and the vertical velocity against time (v_y vs. t) for the large ball, using the same process you did for the small ball. what can you say about the vertical displacement and velocity of the large ball?

Explanation:

When a ball is dropped (assuming free - fall with negligible air resistance for the large ball, similar to the small ball), the vertical displacement \(y\) as a function of time \(t\) follows the equation \(y = y_0-\frac{1}{2}gt^{2}\) (where \(y_0\) is the initial height and \(g\) is the acceleration due to gravity). So the \(y\) vs. \(t\) graph should be a downward - opening parabola (quadratic curve) because the displacement is a quadratic function of time. For the vertical velocity \(v_y\) as a function of time, the equation is \(v_y=v_{0y}-gt\) (with \(v_{0y} = 0\) for a dropped ball), so the \(v_y\) vs. \(t\) graph should be a straight line with a negative slope (since velocity is increasing in the downward direction, and the slope is \(-g\)). The vertical displacement increases non - linearly (quadratically) with time, and the vertical velocity increases linearly with time (in the downward direction, assuming downward is negative or positive depending on the coordinate system) as the ball accelerates due to gravity.

Answer:

For the large ball (in free - fall, neglecting air resistance): The vertical displacement (\(y\) vs. \(t\)) graph is a quadratic (parabolic) curve (since \(y\) depends on \(t^{2}\) as \(y=y_0 - \frac{1}{2}gt^{2}\)), showing that displacement increases non - linearly with time. The vertical velocity (\(v_y\) vs. \(t\)) graph is a straight line with a negative slope (from \(v_y=v_{0y}-gt\), with \(v_{0y} = 0\) for a dropped ball), meaning velocity increases linearly with time (in the downward direction) as the ball accelerates due to gravity.