Question

due oct 7 by 10am points 10 submitting on paper available sep 30 at 12am - oct 28 at 11:59pm exit ticket solve. a diver jumps from a platform. their height in meters after t seconds is given by: h(t)=-4.9t² + 3t + 10. identify the vertex (use formula). interpret what it means. find when they will hit the water (solve h(t)=0)

Explanation:

Step1: Find the x - coordinate of the vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is given by $t=-\frac{b}{2a}$. Here, $a=-4.9$, $b = 3$, so $t=-\frac{3}{2\times(-4.9)}=\frac{3}{9.8}\approx0.31$.

Step2: Find the y - coordinate of the vertex

Substitute $t = \frac{3}{9.8}$ into $h(t)=-4.9t^{2}+3t + 10$.
$h(\frac{3}{9.8})=-4.9\times(\frac{3}{9.8})^{2}+3\times\frac{3}{9.8}+10$
$=-4.9\times\frac{9}{96.04}+\frac{9}{9.8}+10$
$=-\frac{44.1}{96.04}+\frac{9}{9.8}+10$
$=- 0.46+0.92 + 10=10.46$. So the vertex is $(\frac{3}{9.8},10.46)\approx(0.31,10.46)$.
The vertex represents the maximum height of the diver and the time at which the maximum height is reached. The x - coordinate ($t\approx0.31$ s) is the time when the diver reaches the maximum height, and the y - coordinate ($h\approx10.46$ m) is the maximum height above the water.

Step3: Find when the diver hits the water

Set $h(t)=0$, so $-4.9t^{2}+3t + 10 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-4.9$, $b = 3$, $c = 10$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(3)^{2}-4\times(-4.9)\times10=9 + 196=205$.
Then $t=\frac{-3\pm\sqrt{205}}{2\times(-4.9)}=\frac{-3\pm14.32}{-9.8}$.
We have two solutions for $t$:
$t_1=\frac{-3 + 14.32}{-9.8}=\frac{11.32}{-9.8}\approx - 1.15$ (rejected since time cannot be negative)
$t_2=\frac{-3-14.32}{-9.8}=\frac{-17.32}{-9.8}\approx1.77$.

Answer:

• Vertex: $(0.31,10.46)$
• Interpretation: The diver reaches a maximum height of approximately $10.46$ m at $t\approx0.31$ s.
• Time when hitting the water: $t\approx1.77$ s