• 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.
