Assuming the particle is in free fall once it is shot up, its vertical velocity <em>v</em> at time <em>t</em> is
<em>v</em> = 30 m/s - <em>g t</em>
where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height <em>y</em> is given by
<em>y</em> = (30 m/s) <em>t</em> - 1/2 <em>g t</em> ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - <em>g t</em>
<em>t</em> = (30 m/s) / <em>g</em> ≈ 3.06 s
at which point the particle's maximum height would be
<em>y</em> = (30 m/s) (3.06 s) - 1/2 <em>g</em> (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, <em>t</em> ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) <em>t</em> - 1/2 <em>g t</em> ²
Solve for <em>t</em> to get a time of about <em>t</em> ≈ 7.1 s
