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I have given a problem to solve for the displacement of an object given the acceleration $-10\;\mathrm{m/s^2}$ and the initial velocity $627\;\mathrm{m/s}$ of how far an object shoots up, though I am not given the time, so I can't use the equation

$$S=vt+\frac{at^2}2$$

Is there a way to solve this? Thanks!

Edit Thanks to whoever edited my question for the formatting

ariagno
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  • How about finding the time when the velocity gets zero with $v(t)=v_0+a t$ (which would be mathematically equivalent to finding the maximum of $s(t)$) ? If you get this time then you can use it in the equation you wrote down in your post to get the distance. – N0va Sep 14 '16 at 22:26
  • Are you allowed calculus? – J126 Sep 14 '16 at 22:38
  • @JoeJohnson126 Yes. – ariagno Sep 14 '16 at 23:10

2 Answers2

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You can use Torricelli's equation: $$v_f^2 = v_0^2+2a\Delta S$$ where $v_f, v_0$ are, respectively, the final and initial velocity of the object, $a$ its acceleration and $\Delta S$ its displacement, and solve for $\Delta S$.

You know that at the point of maximum height, the object's velocity is 0 $m/s$. Substituting the other terms, and solving for $\Delta S$, we have $$\Delta S = \frac{-v_0^2}{2a} = \frac{-627^2}{-20} = 19 656.45 m$$

Vitor Borges
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  • How did you get the final velocity in that equation? – ariagno Sep 15 '16 at 00:27
  • You want to find the object's maximum height, but this is the moment right before it starts to fall. Falling means its velocity is negative. Which means that at the point of maximum height its velocity is zero, and therefore $v_f = 0$. – Vitor Borges Sep 15 '16 at 00:38
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If acceleration is $-10$, and velocity is an antiderivative of acceleration, then $$ v(t) = \int -10 \, dt = -10t + C. $$ If you put in $t = 0$ you will see that $C$ is your initial velocity. Then you have that displacement is an antiderivative of velocity.

J126
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