Question

A ball is thrown upward at a velocity of 19.6 m/s. What is its velocity after 3.0 s, assuming negligible air resistance?

Answer 1

The ball will be travelling downward with a speed of 9.8 m/s.

Explanation:

The acceleration of the ball will be a constant `9.8m/s^2` (downward), regardless of whether the ball is travelling upward or downward.

The equation that makes this calculation simplest is

`v_f =v_i-g(Deltat)`

where the negative sign has been usedd to account for hte opposite directions of `v_i` and `g`. After 3 seconds,

`v_f = 19.6-(9.8)(3) = 19.6-29.4 = -9.8 m/s`

The negative result tells us that the ball is on its way downward by this time (consistent with our choice of a negative value for the downward acceleration of gravity).