A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of g = 10 m/s2, what?

Question

A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of g = 10 m/s2, what are the magnitude and direction of the ball's velocity at the following times?

1.1 s after it is thrown

Answer 1

The ball is constantly accelerating downward at a rate of 10 m/s^2 (due to gravity).
The change in velocity of the object undergoing uniform acceleration is:
(delta v) = a * t
Where (delta v) is the change in velocity, a is the constant acceleration, and t is the length of time the object is accelerating.

The overall velocity of an object is given as,
V = v_0 + (delta v)
Where v_0 is the initial velocity

For this problem, let’s call the upward direction "positive" and the downward direction "negative".
So now we have,
v_0 = 13 m/s
a = -10 m/s^2
t = 1.1 second

V = (13 m/s) + (-10 m/s^2) * (1.1 seconds)
V = 13 m/s - 11 m/s
V = 2 m/s

Because the sign of the answer is positive, the ball is still traveling in the upward direction.

After 1.1 seconds, the ball is traveling upward at a speed of 2 m/s.