A ball is thrown upward and has a velocity of 67.8 ft/s after 1.0s. After 3.0s, the ball's velocity is 3.4 ft/s. Find the acceleration (in ft/s2) and initial velocity (in ft/s) of the ball. Complete the equation of motion for the velocity of the ball:

v(t) = t +

in 2 seconds velocity decreases by 64.4 ft/s

a = change in v/change in time
= -64.4/2 = -32.2 ft/s^2
(which by the way is about g on earth)
v = 67.8 - 32.2 (t-1)
v = 100 -32.2 t

To find the acceleration and initial velocity of the ball, we can use the equations of motion. The equation of motion for the velocity of an object is:

v(t) = v0 + at

Where v(t) is the velocity at time t, v0 is the initial velocity, a is the acceleration, and t is the time.

Given that the velocity of the ball after 1.0s is 67.8 ft/s and the velocity after 3.0s is 3.4 ft/s, we can plug these values into the equation of motion to get two equations:

67.8 = v0 + 1.0a -- equation 1
3.4 = v0 + 3.0a -- equation 2

Now we can solve these two equations simultaneously to find the values of a and v0.

First, we'll subtract equation 2 from equation 1:

67.8 - 3.4 = (v0 + 1.0a) - (v0 + 3.0a)
64.4 = v0 + a - v0 - 3a
64.4 = -2a

Now we can solve for a:

a = -64.4/2
a = -32.2 ft/s^2

Substituting this value of a into equation 1, we can solve for v0:

67.8 = v0 + 1.0(-32.2)
67.8 = v0 - 32.2
v0 = 67.8 + 32.2
v0 = 100 ft/s

Therefore, the acceleration of the ball is -32.2 ft/s^2 and the initial velocity is 100 ft/s.

Now, we can complete the equation of motion for the velocity of the ball:

v(t) = t + 100 - 32.2t
v(t) = 100 - 31.2t