a basketball rolls off a deck that is 3.2 m from the pavement. the basketball lands 0.75 m from the edge of…

a basketball rolls off a deck that is 3.2 m from the pavement. the basketball lands 0.75 m from the edge of the deck. how fast was the basketball rolling? seema lists the given values in a chart and determines that the unknown value is $v_x$.\n| given | value |\n| ---- | ---- |\n| $delta y$ | 3.2 m |\n| $a_y$ | -9.8 m/s² |\n| $delta x$ | 0.75 m |\nwhich describes seemas error?\nthe $delta y$ and $delta x$ values are switched.\nthe unknown is $\theta$, not $v_x$.\nthe value for $a_y$ is not known.\nthe value of $delta y$ should be -3.2 m.

a basketball rolls off a deck that is 3.2 m from the pavement. the basketball lands 0.75 m from the edge of the deck. how fast was the basketball rolling? seema lists the given values in a chart and determines that the unknown value is $v_x$.\n| given | value |\n| ---- | ---- |\n| $delta y$ | 3.2 m |\n| $a_y$ | -9.8 m/s² |\n| $delta x$ | 0.75 m |\nwhich describes seemas error?\nthe $delta y$ and $delta x$ values are switched.\nthe unknown is $\theta$, not $v_x$.\nthe value for $a_y$ is not known.\nthe value of $delta y$ should be -3.2 m.

Answer

Explanation:

Step1: Analyze vertical - motion

The basketball is in free - fall in the vertical direction. The initial vertical velocity (v_{0y}=0\ m/s), the vertical displacement (\Delta y=- 3.2\ m) (downward is negative), and the acceleration (a_y=-9.8\ m/s^{2}). Using the equation (\Delta y = v_{0y}t+\frac{1}{2}a_yt^{2}), since (v_{0y} = 0\ m/s), we have (\Delta y=\frac{1}{2}a_yt^{2}).

Step2: Solve for time (t)

From (\Delta y=\frac{1}{2}a_yt^{2}), we can re - arrange for (t): (t=\sqrt{\frac{2\Delta y}{a_y}}). Substituting (\Delta y=-3.2\ m) and (a_y = - 9.8\ m/s^{2}), we get (t=\sqrt{\frac{2\times(-3.2)}{-9.8}}\ s\approx\sqrt{\frac{6.4}{9.8}}\ s\approx0.81\ s).

Step3: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ((a_x = 0\ m/s^{2})), and the horizontal displacement (\Delta x = 0.75\ m). The horizontal velocity (v_x) is constant, and we use the equation (\Delta x=v_x t). Then (v_x=\frac{\Delta x}{t}). Substituting (\Delta x = 0.75\ m) and (t\approx0.81\ s), we get (v_x=\frac{0.75}{0.81}\ m/s\approx0.93\ m/s).

Now, for Seema's error: The basketball is falling downwards, so the vertical displacement (\Delta y) should be negative. Seema listed (\Delta y = 3.2\ m) instead of (\Delta y=-3.2\ m).

Answer:

The value of (\Delta y) should be (-3.2\ m).