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This must a simple math problem, but i'm scratching my head here.

An object is falling from its resting position with constant acceleration $9.8m/s^2$ (gravity) and hits the ground with velocity $29.4m/s$. I need to know from what distance this object fell.

I know that "flight-plan" was like this:

$$\begin{array}{c|c|c|} \text{Time ($s$)} & \text{Acceleration ($m/s^2$)} & \text{Velocity ($m/s$)} \\ \hline \text{1} & 9.8 & 9.8 \\ \hline \text{2} & 9.8 & 19.6 \\ \hline \text{3} & 9.8 & 29.4 \\ \hline \end{array}$$

I can establish time traveled by dividing velocity by acceleration: $29.4 / 9.8 = 3s$. To check the units: $(m/s) / (m/s^2) = (m / s * s^2) / (m) = (m *s) / (m) = s$ - seems okay. Now i need to transform time and acceleration into distance. Can i do that?

There are some pages on the Internet that suggest this formula: $$s=v_0t + 1/2at^2$$

$s$ - distance, $v_0$ - initial velocity, $t$ - time, $a$ - acceleration. In this case $v_0$ is zero, so it can be simplified to $$s=1/2at^2$$

But this can't be right, because if i plug values into it when $t = 1$, then i get $1/2 * 9.8 * 1^2 = 1/2 * 9.8 = 4.9m$, but it has to be $9.8m$ if time is $1s$ and acceleration is $9.8m/s^2$. No? If i put other values, then i also don't get result that i expect.

Is there an actual formula to get traveled distance from time and constant acceleration?

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3 Answers 3

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You're going wrong in saying that the body must cover $9.8 m$. The body only gains a velocity of $9.8 m/s$ over the course of $1 s$. Because the body gains a velocity of $9.8 m/s$ only just after one second it should be clear to you that the distance covered in $1s$ must be less than $9.8m$ and so is the case as you have verified

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  • $\begingroup$ Indeed, and since we have uniform acceleration from 0m/s to 9.8m/s, the average velocity is just the average of those two values (4.9m/s). The distance covered is the average velocity (4.9m/s) times the duration (1s), for a result of 4.9m. $\endgroup$ Apr 28, 2020 at 20:06
  • $\begingroup$ @NuclearWang True that! $\endgroup$
    – sai-kartik
    Apr 28, 2020 at 20:10
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The acceleration is the change in velocity.

During the first second, the velocity increases from zero to $9.8$ m/s.

The distance traveled during the first second, therefore, must be less than $9.8$ m.

The formula you have is correct.

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  • $\begingroup$ Thank you for explaining this. Indeed, at t=0 velocity was zero and at t=1 velocity was 9.8m/s^2, therefore average velocity during that time was 4.9m/s and object traveled 4.9m during first second. $\endgroup$
    – plasmatron
    Apr 28, 2020 at 20:13
  • $\begingroup$ @plasmatron exactly. $\endgroup$
    – John
    Apr 28, 2020 at 21:03
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No, it does not have to be $9.8$ m. Here, $u=0, v=29.4, a=9.8$

Using $v^2 - u^2 = 2aS$ we get $$(29.4)^2-0=2\cdot 9.8 \cdot S$$ or $$S=44.1 m$$

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