Please show solution... freefall problem...?

Question

Veronica angrily throws her engagement ring straight up from the roof of a building, 12.0m above the ground, with an initial speed of 6.00m/s. Air resistance may be ignored. For the motion from her hand to the ground, what are the magnitude and direction of a.) the average velocity of the ring? b) the average acceleration of the ring? (P.S. the answer for a .) is -5.23 m/s and for b.) is -9.8m/s^2.... i need the solution... thanks and more power!)

Answer 1

(a)

Taking ground level as the origin and measuring positive distance upwards, the time t to reach ground is t where:

0 = 12.0 + 6.00t - 9.81t^2 / 2

The positive root is:

t = [ - 6 - sqrt(6^2 + 2 * 9.81 * 12) ] / [ - 9.81 ]

= 2.291 sec.

Average velocity is total signed distance (- 12.0m) divided by total time

= - 12 / 2.291

= - 5.24 m / sec.

= 5.24 m / sec. downwards.

(b)

The acceleration due to gravity is - 9.81 m/s^2 upwards or 9.81 m/s^2 downwards, and as this is constant, this is also its average value.