To find when the jumper hits the river, set 

ℎ

(

𝑡

)

=

0

h(t)=0 and solve for 

𝑡

t:

−

0.5

𝑡

2

+

210

=

0

−0.5t 

2

 +210=0

Solve for 

𝑡

2

t 

2

 :

−

0.5

𝑡

2

=

−

210

⇒

𝑡

2

=

210

0.5

=

420

−0.5t 

2

 =−210⇒t 

2

 = 

0.5

210

​

 =420

Take the square root:

𝑡

=

420

≈

20.49
 
seconds

.

t= 

420

​

 ≈20.49seconds.

So, the bungee jumper touches the river at around 20.49 seconds

Let's go step-by-step to clarify the solution.

You are solving for when the bungee jumper hits the river, represented by \( h(t) = 0 \), where the height function is given as:

\[
h(t) = -0.5t^2 + 210
\]

### Step 1: Set \( h(t) = 0 \)
\[
-0.5t^2 + 210 = 0
\]

### Step 2: Isolate the quadratic term
Move the constant to the other side:

\[
-0.5t^2 = -210
\]

### Step 3: Divide both sides by -0.5
\[
t^2 = \frac{210}{0.5} = 420
\]

### Step 4: Take the square root
\[
t = \sqrt{420} \approx 20.49 \, \text{seconds}
\]

Thus, the jumper hits the river at approximately **20.49 seconds**.

Would you like more details or have any questions about this solution?

Here are 5 related questions:
1. What would the equation look like if the jumper started at a different height?
2. How would the result change if gravity (9.8 m/s²) were altered?
3. What would happen if air resistance was included in the model?
4. Could we solve this equation using a different method, such as factoring?
5. How would we handle this if the equation were cubic instead of quadratic?

**Tip:** Always remember to check if the solutions make physical sense—negative time, for example, would not be meaningful in this context!

To find when the jumper hits the river, set h(t)=0 and solve for t: -0.5t^2 + 210 = 0. Solve for t^2: -0.5t^2 = -210 ⇒ t^2 = 210/0.5 = 420. Take the square root: t = sqrt(420) ≈ 20.49 seconds. So, the bungee jumper touches the river at around 20.49 seconds.

In this problem, we solve a quadratic equation to determine when a bungee jumper touches the river. Using concepts from algebra and kinematics, we apply the quadratic formula and solve for time. This solution is suitable for students in Grades 9-11, covering key topics such as quadratic equations and motion under gravity.

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