Question

An object is thrown vertically upward such that it has a speed 25m/s when it reaches two thirds of its maximum height above the launch point

Determine this maximum height.

Answer 1

Step-by-step explanation:

dropped from maximum height were vertical velocity is zero, the object would have a velocity of 25 m/s in a distance of

v² = u² + 2as

s = (v² - u²) / 2a

s = (25² - 0²) / (2(9.8))

s = 31.887755... m

s/h = 1 - ⅔

s/h = ⅓

h = 3s

h = 3(31.887755)

h = 95.6632...

h = 96 m