Solving h=1/2gt^2: Questions Explained

[atypical]

So h=1/2gt^2. I am having some trouble understanding this equation.
Here are my questions:
1. I know you get two completely different answers if you plug in a set value, but why can't you measure height of a free-falling object by multiplying the number of seconds^2 times 9.8m/s^2?
2. Is the coefficient 1/2 because you are compensating for the seconds^2?
3. Does the equation h=1/2gt^2 result because acceleration is the second derivative of position?
4. Why?

 

[quenderin]

1. So you're saying h = gt^2 instead of 1/2gt^2? Well, you start with constant acceleration, g. Starting from rest, this means your velocity is gt. How do you get the distance traveled from the velocity? IF the velocity were constant, you would just take velocity * time. But your velocity is changing over time, so at earlier stages, 1 second of traveling doesn't cover as much ground as it does later. The right way to find the height would be perform the integral

$$\int v \,dt = \int gt \,dt = \frac{1}{2}gt^{2}$$

So that's where the 1/2 comes from. It turns out that the coefficient of 1/2 rightly compensates for the fact that at the beginning you aren't moving as fast as you are at the end.

 

[zgozvrm]

1. I know you get two completely different answers if you plug in a set value

Really? How is that? What "set value" are you plugging in that gives you "two completely different answers?"

1. ...why can't you measure height of a free-falling object by multiplying the number of seconds^2

The complete equation is

$$D = V_it + \frac{1}{2}at^2$$


In the case of horizontal movement of a projectile, there is no acceleration, so we have

$$D_x = V_{ix}t$$


In the case of vertical movement of a projectile, the acceleration is equal to g (which is approximately -32 ft/sec or -9.8 m/s). Note that the acceleration is negative, since it is a downward acceleration. The equation for vertical movement is then

$$D_y = V_{iy}t + \frac{1}{2}gt^2$$


Also, note that [itex]V_{ix}[/tex] and [itex]V_{iy}[/tex] refer to <i>initial</i> horizontal and vertical speeds, respectively.<br /> <br /> The equation you gave only works for an objects with an initial vertical velocity of zero (at rest), in which case [itex]V_{iy} = 0[/tex]<br /> <br /> So, you CAN measure (determine) the height of an object with this equation, as long as you know [itex]V_{iy}[/tex], t, and a.<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="atypical" data-source="post: 2968896" cite="https://www.physicsforums.com/goto/post?id=2968896" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> atypical said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 2. Is the coefficient 1/2 because you are compensating for the seconds^2?<br /> 3. Does the equation h=1/2gt^2 result because acceleration is the second derivative of position? </div> </div> </blockquote>This results from combining the 2 basic forms of the kinematic equations:<br /> <br /> $$D_f = D_i + \frac{V_i + V_f}{2} \, t$$<br /> <br /> and<br /> <br /> $$V_f = V_i + at$$<br /> <br /> <br /> These equations assume constant acceleration.<br /> <br /> The first one says that the final displacement [itex]D_f[/tex] is equal to the initial displacement [itex]D_i[/tex] plus the average velocity over the distance traveled times the time it took to cover that distance.<br /> <br /> The second one says that the final velocity [itex]V_f[/tex] is equal to the initial velocity [itex]V_i[/tex] plus the product of the acceleration and the time it accelerated.<br /> <br /> Do some algebra, and you'll see that you come up with the equation I listed in the first answer.<br /> <br /> (Note that the 1/2 term comes from the average velocity in the first equation, above.)<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="atypical" data-source="post: 2968896" cite="https://www.physicsforums.com/goto/post?id=2968896" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> atypical said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 4. Why? </div> </div> </blockquote>Why what? What is your question here?[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]