Flight Path of a BirdSome birds, like parrots, have strong, large beaks to break open nuts and shells. But other birds, like crows, crack open their food by dropping it to the ground. How can a bird change its flight path to protect its falling food from other hungry birds?Walnuts must fall at least 12 feet to the ground to break open. This creates a fall time that leaves the walnut vulnerable to theft by other birds. Scientists have studied crows dropping their food, and have concluded that the birds actually strategize about where to drop their food. A crow optimizes its path to find the shortest fall time that will return a broken nut.1. The equation y=h0 -16t^2 models the height y of a walnut based on the initial height h0 of the walnut, and the number of seconds t after the walnut is dropped. Write an equation representing the position of the walnut when a crow is flying and carrying the walnut at a height of 25 feet.2. Insert a graph of your equation with the axes labeled. What does the y-axis represent? What does the x-axis represent? Are negative values of x meaningful? Explain your reasoning. What is the meaning of the y-intercept? What is the meaning of the positive x-intercept?3. Using your graph, will it take less than 1 second, 1 second exactly, or more than 1 second for the walnut to hit the ground?4. How can you use your equation to calculate the exact time at which the walnut will hit the ground?5. What do you notice about your binomial? How can this help you solve your equation? When will the walnut hit the ground?

Answer:Equation of walnut dropped as function of time: height [tex]y=25-16t^2[/tex]And the time it takes to hit the ground is: t= 1\frac{1}{4} \,secFor all other details asked, please look at the attached image and the explanations below.Step-by-step explanation:1. If the equation [tex]y=h_0-16\,t^2[/tex] represents the height of the walnut being dropped from height [tex]h_0[/tex], then when the walnut is dropped from a height of 25 feet would be (replacing [tex]h_0[/tex] with 25): [tex]y=25-16\,t^2[/tex] 2. Please see the graph of this quadratic equation in the attached image.a) The y axis represents the height of the walnut (in feet) as it falls for different times (which are represented by the horizontal axis). The horizontal axis represents then the elapsed time for the walnut in its descending path.b) Negative values of the horizontal axis are NOT meaningful since they would represent NEGATIVE times  and not describe the downward motion from the instant (time zero) the bird releases the walnut. c) The y-intercept indicates the height of the walnut at time zero (when the bird releases it).d) The positive x-intercept (where the graph crosses the horizontal axis) represents the time (counted from time zero of the release) that it takes the walnut to reach height 0 feet (that is to hit the ground)3. We can see from the graph that the crossing of the horizontal axis takes more than one second. This is marked with a red dot, and one can see that the crossing takes place to the right of the tick marked as 1 second, meaning that it is at a value greater than 1 second.4. We can use the equation by solving for the unknown "t" when the height of the walnut is zero (y = 0) meaning it reached the ground:[tex]y=25-16\,t^2\\0=25-16\,t^2\\16\,t^2=25\\t^2=\frac{25}{16}\\t=+/-\sqrt{\frac{25}{16} } \\t=+/-\frac{5}{4} \,sec[/tex]5. The equation can also be solved by observing that the binomial that equals zero is actually a difference of squares which can be easily factored out. notice that 25 is the same as [tex]5^2[/tex], 16 is [tex]4^2[/tex], and t is also to the square power ([tex]t^2[/tex]):[tex]0=25-16\,t^2\\0=5^2-4^2\,t^2\\0=5^2-(4t)^2\\0=(5-4t)(5+4t)\\[/tex]This product of binomials can be zero if either factor renders zero. That is:if [tex](5-4\,t)=0\\5=4\,t\\t=\frac{5}{4}[/tex] secondsor if [tex](5+4\,t)=0\\5=-4\,t\\t=-\frac{5}{4}[/tex] secondsThis second result is not something we can use because it considers NEGATIVE times which have no meaning for our problem (times before the walnut was dropped), so we keep only the first result :[tex]t=\frac{5}{4} \,sec = 1\frac{1}{4} \,sec[/tex]