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Two balls are thrown upward from the edge of the cliff in Example 7. The first is thrown with a speed of $ 48\;ft/s $ and the other is thrown a second later with a speed of $ 24\;ft/s $. Do the balls ever pass each other?

The two balls pass each other at $t=5 \mathrm{s}$

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So we know, um, that our position function is written a Z s equals one half a t squared, plus the initial velocity times t plus the initial position. So we're gonna be substituting a being equal to negative 32 and the initial being 48. So now we have s one of t is equal to negative 16 t squared plus 48 t plus s not. And then, um, if we have an initial velocity of 24 so it will be our second position function will have a negative 16 t squared plus 24 t plus, that's not so. Since we have this position functions, we want to know if the two balls will pass each other. Um, in other words, we know that the two balls will pass each other if S one of t is equal to s two of tea minus one. So based on that, what we're gonna have is negative. 16 t squared plus 48 t plus s not is equal to negative. 16 times T minus one squared, plus 24 times T minus one plus s. Not so because the S knots on both sides, they cancel so that the initial position ultimately doesn't matter. We foil everything out. Combine like terms, get it all into one side. Um, and what we end up getting you're just combining these things is negative. 80 equals negative. 40. So that means T equals five. So we see that the two balls will pass each other at T equals five seconds.

California Baptist University