• The only force acting on the ball, and thus the net force, is due to its own weight, with magnitude

<em>w</em> = <em>m</em> <em>g</em> = (0.4 kg) (9.8 m/s²) ≈ **3.9 N**

pointing downward.

• The ball is in free fall, so its acceleration is <em>g</em> = **9.8 m/s²** (also pointing downward).

• The ball is dropped from a height of 49 m, so its height <em>y</em> at time <em>t</em> is

<em>y</em> = 49 m - 1/2 <em>g</em> <em>t</em> ²

Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 49 m - 1/2 <em>g t</em> ²

<em>t</em> ² = (98 m) / <em>g</em>

<em>t</em> = √((98 m) / <em>g</em>) = √(10) s ≈ **3.2 s**

• The ball's velocity <em>v</em> at time <em>t</em> is

<em>v</em> = - <em>g t</em>

so that at the time found previously, the ball will have attained a velocity of

<em>v</em> = - <em>g</em> (3.2 s) ≈ -31 m/s

and thus a <em>speed</em> of about **31 m/s**.