Question Statement

Suppose a ball is thrown into the air and its height $h (t)$ after $t$ seconds is given by the parabolic trajectory: $h (t) = - 6 t^{2} + 10 t + 5$

If this ball hits a wall representing the equation: $h (t) = 9 t$

By drawing the graphs, find out when and where the ball reaches the wall.

Background and Explanation

To find where two trajectories meet, we analyze the intersection of a quadratic function (the parabola) and a linear function (the wall). Graphing these functions requires identifying key points such as intercepts and the vertex, while the exact meeting point is found by setting the two equations equal to each other.

Solution

To solve this problem, we first determine the key features of the parabolic path to assist in graphing, and then calculate the point of intersection with the wall.

Step 1: Find the Intercepts of the Parabola

To graph $h (t) = - 6 t^{2} + 10 t + 5$ , we find where the ball would theoretically hit the ground ( $x$ -intercepts) and its starting height ( $y$ -intercept).

For the $x$ -intercept, set $h (t) = 0$ : $- 6 t^{2} + 10 t + 5 = 0$

Using the quadratic formula $t = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ : $t = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 6 ) ( 5 )}{2 ( - 6 )}$ $t = \frac{- 10 \pm 100 + 120}{- 12}$ $t = \frac{- 10 \pm 220}{- 12} \approx \frac{- 10 \pm 14.83}{- 12}$

This gives two possible values for $t$ :

$t = \frac{- 10 + 14.83}{- 12} \approx - 0.4025$ (Ignore as time cannot be negative)
$t = \frac{- 10 - 14.83}{- 12} \approx 2.069$

For the $y$ -intercept, set $t = 0$ : $h (0) = - 6 (0)^{2} + 10 (0) + 5 = 5$ The ball is thrown from an initial height of $5$ meters.

Step 2: Find the Intersection with the Wall

The ball reaches the wall when the height of the trajectory equals the height of the wall equation. We set the two equations for $h (t)$ equal to each other: $- 6 t^{2} + 10 t + 5 = 9 t$

Rearrange the equation into standard quadratic form: $- 6 t^{2} + t + 5 = 0$

Multiply by $- 1$ to simplify: $6 t^{2} - t - 5 = 0$

Factor the quadratic: $(6 t + 5) (t - 1) = 0$

Solving for $t$ :

$6 t + 5 = 0 \Rightarrow t = - 5/6$ (Ignore negative time)
$t - 1 = 0 \Rightarrow t = 1$

Step 3: Determine the Height at Impact

Substitute $t = 1$ back into the wall equation to find the height: $h (1) = 9 (1) = 9$

Conclusion: The ball reaches the wall at $t = 1$ second at a height of $9$ meters.

Key Formulas or Methods Used

Quadratic Formula: $t = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Intercept Method: Setting $h (t) = 0$ for horizontal intercepts and $t = 0$ for vertical intercepts.
System of Equations: Setting two functions equal to find their point of intersection.

Summary of Steps

Identify the two equations: the parabola $h (t) = - 6 t^{2} + 10 t + 5$ and the line $h (t) = 9 t$ .
Calculate the $x$ -intercepts of the parabola using the quadratic formula to help define the graph's boundaries.
Calculate the $y$ -intercept of the parabola to find the starting height.
Set the two equations equal to each other to find the time ( $t$ ) of impact.
Solve the resulting quadratic equation for $t$ .
Plug the value of $t$ back into either original equation to find the height ( $h$ ) of impact.

Background and Explanation

Step 1: Find the Intercepts of the Parabola

To graph

h (t) = - 6 t^{2} + 10 t + 5

, we find where the ball would theoretically hit the ground (

x

-intercepts) and its starting height (

y

-intercept).

For the $x$ -intercept, set $h (t) = 0$ :

- 6 t^{2} + 10 t + 5 = 0

Using the quadratic formula

t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 6 ) ( 5 )}{2 ( - 6 )}

t = \frac{- 10 \pm 100 + 120}{- 12}

t = \frac{- 10 \pm 220}{- 12} \approx \frac{- 10 \pm 14.83}{- 12}

This gives two possible values for

t

t = \frac{- 10 + 14.83}{- 12} \approx - 0.4025

(Ignore as time cannot be negative)

t = \frac{- 10 - 14.83}{- 12} \approx 2.069

For the $y$ -intercept, set $t = 0$ :

h (0) = - 6 (0)^{2} + 10 (0) + 5 = 5

The ball is thrown from an initial height of

5

meters.

Step 2: Find the Intersection with the Wall

The ball reaches the wall when the height of the trajectory equals the height of the wall equation. We set the two equations for

h (t)

equal to each other:

- 6 t^{2} + 10 t + 5 = 9 t

Rearrange the equation into standard quadratic form:

- 6 t^{2} + t + 5 = 0

Multiply by

- 1

to simplify:

6 t^{2} - t - 5 = 0

Factor the quadratic:

(6 t + 5) (t - 1) = 0

Solving for

t

6 t + 5 = 0 \Rightarrow t = - 5/6

(Ignore negative time)

t - 1 = 0 \Rightarrow t = 1

Summary of Steps

Identify the two equations: the parabola

h (t) = - 6 t^{2} + 10 t + 5

and the line

h (t) = 9 t

Calculate the

x

-intercepts of the parabola using the quadratic formula to help define the graph's boundaries.

Calculate the

y

-intercept of the parabola to find the starting height.

Set the two equations equal to each other to find the time (

t

) of impact.

Solve the resulting quadratic equation for

t

Plug the value of

t

back into either original equation to find the height (

h

) of impact.

1.2 Q-10

Question Statement

Background and Explanation

Solution

Step 1: Find the Intercepts of the Parabola

Step 2: Find the Intersection with the Wall

Step 3: Determine the Height at Impact

Key Formulas or Methods Used

Summary of Steps

1.2 Q-10

Question Statement

Background and Explanation

Solution

Step 1: Find the Intercepts of the Parabola

Step 2: Find the Intersection with the Wall

Step 3: Determine the Height at Impact

Key Formulas or Methods Used

Summary of Steps