Question

1. an osprey, a fish - eating bird of prey, dives towards the water to a salmon. the height (h(t)), in meters, of the osprey above the water (t) seconds after it begins its dive can be approximated by the function: (h(t)=5t^{2}-30t + 45). a) what is the initial height of the osprey? b) when will the osprey catch the fish? c) when is the osprey 34 feet above the water? d) how high was the osprey at 1 second? 4 seconds?

Explanation:

Step1: Find initial height (a)

Set \(t = 0\) in \(h(t)=5t^{2}-30t + 45\).

$$h(0)=5\times0^{2}-30\times0 + 45=45$$

Step2: Find when osprey catches fish (b)

The osprey catches the fish when \(h(t)=0\). So solve the quadratic equation \(5t^{2}-30t + 45 = 0\). First, divide by 5: \(t^{2}-6t + 9=0\). Factor it as \((t - 3)^{2}=0\), then \(t=3\).

Step3: Find when \(h(t)=34\) (c)

Set \(h(t)=34\), so \(5t^{2}-30t + 45=34\). Rearrange to \(5t^{2}-30t+11 = 0\). Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 5\), \(b=-30\), \(c = 11\).

$$t=\frac{30\pm\sqrt{(-30)^{2}-4\times5\times11}}{2\times5}=\frac{30\pm\sqrt{900 - 220}}{10}=\frac{30\pm\sqrt{680}}{10}=\frac{30\pm2\sqrt{170}}{10}=3\pm\frac{\sqrt{170}}{5}$$

Step4: Find height at \(t = 1\) and \(t = 4\) (d)

For \(t = 1\): \(h(1)=5\times1^{2}-30\times1 + 45=5-30 + 45=20\).
For \(t = 4\): \(h(4)=5\times4^{2}-30\times4 + 45=5\times16-120 + 45=80-120 + 45=5\).

Answer:

a) 45 meters
b) 3 seconds
c) \(t = 3+\frac{\sqrt{170}}{5}\) seconds and \(t = 3-\frac{\sqrt{170}}{5}\) seconds
d) 20 meters at 1 second, 5 meters at 4 seconds