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

Question

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.

@Jihad But what is the formula to solve such problem?$s=v_0+(1/2)gt^2$? — Flip, Oct 24 '14 at 17:13
@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
Of course, you should consider the sign of $g$, a correct form should be vector-wise: $S = \vec{v_0}t + \frac{\vec{g}t^2}{2}$ — Andrei Rykhalski, Oct 24 '14 at 17:18
@AndreiRykhalski here I have to use $g=-32ft/sec^2$ or $g=32ft/sec^2$ — Flip, Oct 24 '14 at 17:18
@Flip As $v_0$ is directed opposite to $g$, you should use $g = -32 ft/sec^2$ — Andrei Rykhalski, Oct 24 '14 at 17:19

score 0 · Answer 1 · answered Oct 24 '14 at 17:23

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!

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

1 Answers1