a spring - loaded toy is launched straight upward from the surface of a table. in case 1, the toy is caught…

a spring - loaded toy is launched straight upward from the surface of a table. in case 1, the toy is caught at the instant that it reaches its highest point and held at rest. in case 2, the toy is allowed to rise, then fall back to the table. in which case is the magnitude of the average velocity of the toys free - fall motion the greatest?\na case 1, because the toy travels for less time\nb case 1, because the toy has a greater displacement\nc case 2, because the toy travels a greater distance\nd case 2, because the toy travels for a longer time

a spring - loaded toy is launched straight upward from the surface of a table. in case 1, the toy is caught at the instant that it reaches its highest point and held at rest. in case 2, the toy is allowed to rise, then fall back to the table. in which case is the magnitude of the average velocity of the toys free - fall motion the greatest?\na case 1, because the toy travels for less time\nb case 1, because the toy has a greater displacement\nc case 2, because the toy travels a greater distance\nd case 2, because the toy travels for a longer time

Answer

Explanation:

Step1: Recall average - velocity formula

The formula for average velocity is $\bar{v}=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the displacement and $\Delta t$ is the time - interval.

Step2: Analyze displacement in each case

In Case 1, the toy is caught at its highest point. The displacement of the free - fall motion is the height $h$ from the table to the highest point. In Case 2, the toy goes up to the highest point and then falls back to the starting position (the table). The displacement of the free - fall motion is also the height $h$ from the highest point to the table. So, the displacements in both cases are the same in magnitude.

Step3: Analyze time in each case

The time of free - fall is given by the kinematic equation $h = v_0t+\frac{1}{2}gt^2$ (taking downwards as positive, and at the start of free - fall $v_0 = 0$), so $h=\frac{1}{2}gt^2$ or $t=\sqrt{\frac{2h}{g}}$. In Case 2, the toy has to fall the same height $h$ as in Case 1, but it also has the time to rise back up to the highest point before starting the free - fall. So, the time of free - fall in Case 2 is greater than in Case 1.

Step4: Analyze average velocity

Since $\bar{v}=\frac{\Delta x}{\Delta t}$ and $\Delta x$ (displacement) is the same for both cases, and $\Delta t_2>\Delta t_1$ (time in Case 2 is greater than time in Case 1), the magnitude of the average velocity in Case 1 is greater because the denominator of the average - velocity formula is smaller. Also, the toy travels for less time in Case 1.

Answer:

A. Case 1, because the toy travels for less time