Answer:
15 yd/s
Step-by-step explanation:
You want to know the rate of change of distance to a firework when it is 8 yards away horizontally, 6 yards high, and its height is increasing at 25 yards per second.
Geometry
The geometry of the problem can be modeled by a 3-4-5 right triangle scaled by a factor of 2 yards. The straight-line distance to the firework is ...
5 × (2 yd) = 10 yards
The vertical distance is given as 6 yards, increasing at 25 yd/s. Then the cosine of the angle from the line of sight to the direction of travel is ...
cos(α) = adjacent/hypotenuse = (6 yd)/(10 yd) = 0.6
Speed made good
The speed in the direction of the line of sight is the speed of the firework, multiplied by the cosine of the angle between its direction and the line of sight:
rate of change of distance away = (25 yd/s) × cos(α) = (25 yd/s)(0.6)
rate of change of distance away = 15 yd/s
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Additional comment
The rate of change can also be found using calculus. The distance away can be expressed using the distance formula, and its derivative found. The value of that derivative at the time of interest will give the rate of change of distance to the firework. This is shown in the second attachment, using a graphing calculator to find the derivative.
