An object is thrown vertically up and attains an upward velocity of 34 m/s when it reaches one fourth of its maximum height above its launch point. what was the initial speed of the object?

39.26 m/s

Explanation:

Let the maximum height at which the object reaches is h.

Let the initial speed f the object is u.

velocity at height h/4 is 34 m/s

acceleration due to gravity = 9.8 m/s^2

For AB part

Use third equation of motion for AB part

here, initial velocity is v = 34 m/s

Final velocity at maximum height, v' = 0

Height = h - h/4 = 3h/4

h = 78.64 m

For OA part

initial velocity = u

final velocity, v = 34 m/s

Height = h / 4 = 78.64 / 4 = 19.66 m

Use third equation of motion

u = 39.26 m/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

inverse v=i*r

nice nice