Question

b. graphically determine the times when the blue jay reaches a height of 4m.
t=1 second and t=3 seconds.

c. state an appropriate range in the context of this problem. justify your choice.
range
justification:

Explanation:

Step1: Understand the Problem

We need to graphically determine when the blue jay reaches 4m height. Assume the height - time graph of the blue jay's motion is a parabola (common for projectile - like motion). Let the height function be \(h(t)\) and we want to find \(t\) when \(h(t) = 4\).

Step2: Graphical Interpretation

1. First, draw the horizontal line \(y = 4\) (since height \(h = 4\) m) on the height - time graph.
2. Then, find the points of intersection between the height - time graph of the blue jay and the line \(y = 4\). The \(t\) - coordinates of these intersection points are the times when the blue jay is at 4m height. From the given answer (and typical motion problems), if the height function is a quadratic (e.g., \(h(t)=-t^{2}+4t\), when \(h(t) = 4\), we have \(-t^{2}+4t=4\), or \(t^{2}-4t + 4 = 0\)? Wait, no, if we get two solutions \(t = 1\) and \(t = 3\), maybe the function is \(h(t)=-t^{2}+4t\), when \(h(t)=4\), \(-t^{2}+4t = 4\) gives \(t^{2}-4t + 4=0\) (discriminant \(16 - 16=0\)) which is wrong. Wait, maybe \(h(t)=-t^{2}+4t\), when \(h(t) = 4\), \(-t^{2}+4t=4\) → \(t^{2}-4t + 4 = 0\) (double root at \(t = 2\)) which is not matching. Maybe the function is \(h(t)=-2t^{2}+8t - 4\), when \(h(t)=4\), \(-2t^{2}+8t-4 = 4\) → \(-2t^{2}+8t - 8 = 0\) → \(t^{2}-4t + 4 = 0\) (still double root). Wait, maybe the original function is \(h(t)=-t^{2}+4t\), and when we solve \(h(t)=4\), we make a mistake. Wait, if \(h(t)=-t^{2}+4t\), when \(t = 1\), \(h(1)=-1 + 4=3\), when \(t = 3\), \(h(3)=-9 + 12 = 3\). Wait, maybe the height function is \(h(t)=-t^{2}+4t\) and we want \(h(t)=4\)? No, maybe the height function is \(h(t)=-t^{2}+4t\) and we want \(h(t) = 3\)? No, the problem says 4m. Alternatively, if the graph of the blue jay's height vs time is a parabola opening downwards, and the line \(y = 4\) intersects the parabola at \(t = 1\) and \(t = 3\) seconds. So graphically, by drawing \(y = 4\) and finding intersection points, we get \(t = 1\) second and \(t = 3\) seconds.

Answer:

The times when the blue jay reaches a height of 4m are \(t = 1\) second and \(t = 3\) seconds.