Answer:
∆y = 95 ft
the vertical distance the shark travels is 95 ft
Step-by-step explanation:
Given;
Initial position y1= 80 ft below the surface of water = -80ft
Final Position y2 = 15 ft above the surface = +15ft
The vertical distance travelled by the shark is;
∆y = y2 - y1 = 15 - (-80) ft
∆y = 15 +80 ft
the 19.50 for each one
