Agolf ball is released from rest from the top of a very tall building. choose a coordinate system whose origin is at the starting point of the ball, and whose y axis points vertically upward. neglecting air resistance, calculate the ve- locity of the ball after 3.04 s. the acceleration of gravity is 9.8 m/s2 . answer in units of m/s.

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: and , First we need to find the height to time iqual to 3.04 s using the formula: and remember that golf ball was released from the rest (Vo= 0 (m/s)) so , we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using solving for Vf, we get: replacing the values given , so we get: Vf = 29.79 (m/s).

Explanation:

Here given that:

Now, we need to find the velocity (v) after the time 3.04 s.

From the given and required data we choose the suitable equation of motion:

Putting the given values in the above equation:

∵the given frame of reference points in the upward directoin so we have negative on the downward side.

i think the answer is a, hope this is right, that would be my best guess.

i guess it’s a and b