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 [tex]\(v_x\)[/tex].

| Given | Value |
|------------|--------------|
| [tex]\(\Delta y\)[/tex] | 3.2 m |
| [tex]\(a_y\)[/tex] | -9.8 m/s[tex]\(^2\)[/tex] |
| [tex]\(\Delta x\)[/tex] | 0.75 m |

Which describes Seema's error?
A. The [tex]\(\Delta y\)[/tex] and [tex]\(\Delta x\)[/tex] values are switched.
B. The unknown is [tex]\(\theta\)[/tex], not [tex]\(v_x\)[/tex].
C. The value for [tex]\(a_y\)[/tex] is not known.
D. The value of [tex]\(\Delta y\)[/tex] should be -3.2 m.



Answer :

To solve this problem thoroughly, let's review the information given:

1. The basketball falls off a deck 3.2 meters high.
2. The horizontal distance it travels before landing is 0.75 meters.
3. The acceleration due to gravity acting on the basketball is -9.8 m/s² (negative because gravity acts downwards).

We need to figure out how Seema's understanding or presentation of this data may be incorrect and which statement best describes the error.

Seema lists the following values in her chart:

[tex]\[ \begin{array}{|l|l|} \hline \text{Given} & \text{Value} \\ \hline \Delta y & 3.2 \, \text{m} \\ \hline a_y & -9.8 \, \text{m/s}^2 \\ \hline \Delta x & 0.75 \, \text{m} \\ \hline \end{array} \][/tex]

We know that:

[tex]\[ \Delta y \text{ represents the vertical displacement} \][/tex]

Because the basketball falls downwards, vertical displacement should be considered negative (i.e., [tex]\(\Delta y = -3.2 \text{ m}\)[/tex]).

Let's now look at the given options:

1. The [tex]$\Delta y$[/tex] and [tex]$\Delta x$[/tex] values are switched: This isn't correct because [tex]$\Delta y$[/tex] refers to vertical displacement and [tex]$\Delta x$[/tex] refers to horizontal displacement, which are correctly listed.

2. The unknown is [tex]$\theta$[/tex], not [tex]$v_x$[/tex]: This is not correct as the problem states that we need to find [tex]$v_x$[/tex], the horizontal velocity, making [tex]$v_x$[/tex] the unknown value.

3. The value for [tex]$a_y$[/tex] is not known: This statement is incorrect because [tex]$a_y$[/tex] is given as the acceleration due to gravity, which is a known constant value of [tex]$-9.8 \, \text{m/s}^2$[/tex].

4. The value of [tex]$\Delta y$[/tex] should be [tex]$-3.2 \text{ m}$[/tex]: This is correct because, as already discussed, the vertical displacement should be negative since the basketball is falling downwards. Hence, [tex]$\Delta y$[/tex] must be -3.2 m.

Therefore, Seema's error is specified by:

[tex]\[ \boxed{\text{The value of } \Delta y \text{ should be } -3.2 \text{ m}.} \][/tex]

This answer accurately describes the error in Seema's listing of the given values.