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An object is tossing upwards with an initial speed of $64 \text{ feet/sec}$.Suppose the gravitational acceleration is $32\text{ feet/sec}^2$. Find the smallest $t$ such that the object reaches the height of $96\text{ feet}$ at time $t$.

My problem: Which formula I have to use to find $t$?

Thanks.

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    $\begingroup$ It seems like object will never reach $96$ feet. $\endgroup$ – Jihad Oct 24 '14 at 17:10
  • $\begingroup$ @Jihad But what is the formula to solve such problem?$s=v_0+(1/2)gt^2$? $\endgroup$ – Flip Oct 24 '14 at 17:13
  • $\begingroup$ $S = S_0 + v_0t + \frac{gt^2}{2}$ $\endgroup$ – Andrei Rykhalski Oct 24 '14 at 17:13
  • $\begingroup$ @AndreiRykhalski what is $S_0$? $\endgroup$ – Flip Oct 24 '14 at 17:14
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    $\begingroup$ @Flip Starting space coordinate, in your case an object is on a ground level, so $S_0 = 0$. $\endgroup$ – Andrei Rykhalski Oct 24 '14 at 17:15
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We have $$ s = ut + \frac{1}{2}at^2 $$
which we can rearrange to get $$ t^2 + 2 \dfrac{2u}{a} - 2 \dfrac{s}{a} = 0 $$ and applying the quadratic equation to that (with correct values of s, u, and a) should get you your answer!

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