To solve this problem, let's break it down into two parts: the hawk's motion and the small bird's motion. We will use kinematic equations to determine when and where the hawk catches the small bird.

Hawk's Motion

• The hawk starts at an altitude of 400 m and dives downward with an initial velocity of $v_{0h} = 0$ m/s.
• It accelerates downward at $a_h = 10 \, \text{m/s}^2$.
• The position of the hawk at any time $t$ can be given by the kinematic equation: $y_h(t) = 400 - \frac{1}{2} a_h t^2$ $y_h(t) = 400 - \frac{1}{2} (10) t^2 = 400 - 5t^2$

Small Bird's Motion

• The small bird starts at an altitude of 40 m and flies upward at a constant velocity of $v_b = 5 \, \text{m/s}$.
• The position of the bird at any time $t$ is: $y_b(t) = 40 + v_b t$ $y_b(t) = 40 + 5t$

When Does the Hawk Catch the Bird?

The hawk catches the bird when their positions are equal, i.e., $y_h(t) = y_b(t)$.

Equating the two equations: $400 - 5t^2 = 40 + 5t$ Simplifying: $400 - 40 = 5t + 5t^2$ $360 = 5t + 5t^2$ Divide through by 5: $72 = t + t^2$ Rearrange into a quadratic equation: $t^2 + t - 72 = 0$

Solving the Quadratic Equation

Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $t^2 + t - 72 = 0$, we have $a = 1$, $b = 1$, and $c = -72$. Substituting these values into the formula: $t = \frac{-1 \pm \sqrt{1^2 - 4(1)(-72)}}{2(1)}$ $t = \frac{-1 \pm \sqrt{1 + 288}}{2}$ $t = \frac{-1 \pm \sqrt{289}}{2}$ $t = \frac{-1 \pm 17}{2}$ This gives two possible values for $t$: $t = \frac{-1 + 17}{2} = 8 \quad \text{(positive time, valid solution)}$ $t = \frac{-1 - 17}{2} = -9 \quad \text{(negative time, not physically meaningful)}$

So, the hawk catches the bird after 8 seconds.

How Far Above the Ground Will They Be?

Now, we can substitute $t = 8$ seconds into either the hawk's or the bird's position equation to find the altitude at which they meet.

Using the bird's equation: $y_b(8) = 40 + 5(8) = 40 + 40 = 80 \, \text{m}$

So, they will meet 80 meters above the ground.

Final Answers

• It takes the hawk 8 seconds to catch the small bird.
• They will be 80 meters above the ground when this happens.

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Do you want more details or have any questions? Here are some related questions to further explore this scenario:

1. What would happen if the hawk had an initial downward velocity?
2. How would the situation change if the bird were accelerating upward instead of flying at a constant speed?
3. If the hawk's acceleration was reduced to 5 m/s², how long would it take to catch the bird?
4. What would the situation look like if the bird started flying downward instead of upward?
5. What is the total distance the hawk travels during the dive?

Tip: When solving kinematic problems, always start by identifying known variables and choosing appropriate kinematic equations based on constant acceleration or velocity conditions.