An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point. What was the initial speed of the object? The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s