You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it is traveling at a speed of 25.0 m/s upward. A) Use the work-energy theorem to find its speed just as it left the ground. What is it?
B) Use the work-energy theorem to find its maximum height. What is it?

30.25 m/s

46.68 m

Explanation:

Work Energy theorem states that

W = ½mv2² - ½mv1²

W = ½m(v2² - v1²)

Net work done by the force = -mgd

Net work done = -m * 9.8 * 14.8

Net work done = -145m

Using the work energy theorem

-145m = ½m(v2² - v1²)

-145m = ½ * m(25² - v1²)

-290m = 625m - v1²m

v1² = 625 + 290

v1² = 915

v1 = √915 = 30.25 m/s

B

-mgd = ½m(v2² - v1²), where v2 = 0, so

-mgd = ½mv1²

Making d the subject of the formula, we have

d = -½mv1²/mg

d = v1²/2g

d = 915/ 2 * 9.8

d = 915 / 19.6

d = 46.68 m

does anyone know or what?