1. you hit a tennis ball with an initial vertical velocity, $v_0$, of 15 m/s. the ball leaves the tennis racket 2 meters above the ground
other side of the net, h(t)?
use $h(t) = -5t^2 + v_0t + h_0$ where
$h(t)$ is the height of a tennis ball being hit
$h_0$ is the initial velocity
$v_0$ is the velocity in meters/second after time (t) in seconds

$t = 15.02$ seconds
$t = 3.13$ seconds
$t = 0.03$ seconds
$t = 0.13$ seconds

Question

1. you hit a tennis ball with an initial vertical velocity, $v_0$, of 15 m/s. the ball leaves the tennis racket 2 meters above the ground
other side of the net, h(t)?
use $h(t) = -5t^2 + v_0t + h_0$ where
$h(t)$ is the height of a tennis ball being hit
$h_0$ is the initial velocity
$v_0$ is the velocity in meters/second after time (t) in seconds

$t = 15.02$ seconds
$t = 3.13$ seconds
$t = 0.03$ seconds
$t = 0.13$ seconds

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Solving Quadratic Equations", "Equation Modeling" ], "new_concepts": [ "Projectile Motion Modeling" ], "current_concepts": [ "Solving Quadratic Equations", "Equation Modeling", "Projectile Motion Modeling" ] } ## Identify given parameters Using the **Equation Modeling** knowledge point $$ \begin{aligned} &v_0 = 15\text{ m/s}\ &h_0 = 2\text{ m}\ &h(t) = -5t^2 + 15t + 2 \end{aligned} $$ ## Set up the equation for hitting the ground Using the **Equation Modeling** knowledge point $$ \begin{aligned} &h(t) = 0\ &-5t^2 + 15t + 2 = 0 \end{aligned} $$ ## Solve the quadratic equation Using the *Solving Quadratic Equations* knowledge point $$ \begin{aligned} &t = \frac{-15 \pm \sqrt{15^2 - 4(-5)(2)}}{2(-5)}\ &t = \frac{-15 \pm \sqrt{225 + 40}}{-10}\ &t = \frac{-15 \pm \sqrt{265}}{-10} \end{aligned} $$ ## Calculate the positive time value Using the *Solving Quadratic Equations* knowledge point $$ \begin{aligned} &\sqrt{265} \approx 16.2788\ &t_1 = \frac{-15 + 16.2788}{-10} \approx -0.13\text{ s}\ &t_2 = \frac{-15 - 16.2788}{-10} \approx 3.13\text{ s} \end{aligned} $$ ## Select the physically meaningful solution Using the *Solving Quadratic Equations* knowledge point $$ \begin{aligned} &t > 0 \implies t \approx 3.13\text{ seconds} \end{aligned} $$

(A) $t = 15.02\text{ seconds}$

(B) $t = 3.13\text{ seconds}$

(C) $t = 0.03\text{ seconds}$

(D) $t = 0.13\text{ seconds}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Solving Quadratic Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Identify given parameters

Set up the equation for hitting the ground

Solve the quadratic equation

Calculate the positive time value

Select the physically meaningful solution

Answer: