Question

A shark is 80 feet below the surface of the water. It swims up and jumps out of the water to a height of 15 feet above the surface. Find the vertical distance the shark travels.

Answer 1

∆y = 95 ft

the vertical distance the shark travels is 95 ft

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

∆y = 95 ft

the vertical distance the shark travels is 95 ft