The ball thrown upward has height <em>y</em> and velocity <em>v</em> at time <em>t</em> given by
<em>y</em>₁ = 3 m + (24 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em> ²
<em>v</em>₁ = 24 m/s - <em>g t</em>
while the ball thrown downward has height and velocity
<em>y</em>₂ = 3 m - (24 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em> ²
<em>v</em>₂ = - 24 m/s - <em>g t</em>
where <em>g</em> = 9.80 m/s².
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(a) At its highest point, the first ball has zero velocity:
24 m/s - <em>g t</em> = 0 → <em>t</em> = (24 m/s)/<em>g</em> ≈ 2.45 s
(b) The first ball reaches a maximum height <em>y</em> [max] such that
0² - (24 m/s)² = -2 <em>g</em> <em>y</em> [max] → <em>y</em> [max] = (24 m/s)²/(2<em>g</em>) ≈ 29.4 m
(c) Solve for <em>t</em> when <em>y</em>₁ = 0 and <em>y</em>₂ = 0, then take the absolute difference between them:
0 = 3 m + (24 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em> ² → <em>t</em> ≈ 5.02 s
0 = 3 m - (24 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em> ² → <em>t</em> ≈ 0.123 s
→ |5.02 s - 0.123 s| ≈ 5.14 s