3. with all my strength i throw a 300g snowball vertically in the air. l notice that the snowball has reached its highest point after 3.1 seconds. a) at what speed did the ball leave my hand (numerically)? b) how high was the snowball at the highest point (numerically)? c) suppose i stood on a 1m high ladder at the time of dropping. how many seconds would the snowball be in the air until it hits the ground (numerically)?

A) The velocity decreases at 9.8 m/s, so it started at 9.8·3.1 = 30.38 m/s if it reaches zero at 3.1 seconds.

b) The average velocity over the interval was half that, 15.19 m/s, so the distance above the release point is (3.1 s)·(15.19 m/s) = 47.089 m

c) The time it takes for an object to fall from height h satisfies the relation
  h = 1/2gt²
Solving for t gives us
  t = √(2h/g)
From a height of 1 meter, it takes about 1/√4.9 ≈ 0.452 seconds to fall to the ground.


With no information indicating otherwise, we assume the snowball is released from the level of your feet.

Yup no bull frogs here man