Find the smallest $t$ such that the object reaches the height of $96\text{ feet}$ at time $t$.

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.

• It seems like object will never reach $96$ feet. – Jihad Oct 24 '14 at 17:10
• @Jihad But what is the formula to solve such problem?$s=v_0+(1/2)gt^2$? – Flip Oct 24 '14 at 17:13
• $S = S_0 + v_0t + \frac{gt^2}{2}$ – Andrei Rykhalski Oct 24 '14 at 17:13
• @AndreiRykhalski what is $S_0$? – Flip Oct 24 '14 at 17:14
• @Flip Starting space coordinate, in your case an object is on a ground level, so $S_0 = 0$. – Andrei Rykhalski Oct 24 '14 at 17:15

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!