Vertical Motion vs. Time/Height Parabola Clarification Just for some clarification, $y(t)=y_0 + v_0t + -4.9t^2$ where ($y_0,v_0$ are initial position and initial velocity) is for projectile motion.  Now if a ball is thrown up vertically, with linear motion, the time/height parabola of the ball would not necessarily be of the same format.  
For example,
A ball is throw upward with an initial velocity of $13.36m/s$, $1m$ from the ground.
The time/height quadratic equation of the ball will not be $h(t)=-4.9t^2 +13.36t +1$ correct?
Thank you for the clarification. 
 A: The equation $y(t) = y_0 + v_0t - 4.9t^2$ gives the height of a projectile launched from an initial height of $y_0$ with an initial vertical velocity $v_0$. This equation is exactly the same whether or not the projectile is launched straight into the air or off at some angle.
However, it is important to note that since $v_0$ the vertical component of the velocity. If the projectile is not launched straight upward, but rather at some angle, then $v_0$ is not the full velocity.
For instance, as in your example, the height of an object thrown straight up with an initial velocity of $13.36 \frac{\text{m}}{\text{s}}$ from a height of $1 \text{m}$ is indeed $h(t) = 1 + 13.36t - 4.9t^2$ since $y_0 = 1$ and $v_0 = 13.36$ (all velocity is in the vertical direction). However, if instead the ball is thrown from the same height with the same velocity, but at an angle of $30^\circ$ from the ground, then now the vertical component of the velocity is half the full velocity: $v_0 = 13.36\sin{30^\circ} = 13.36/2 = 6.68$. So now the height is given by $h(t) = 1 + 6.68t - 4.9t^2$.
The equation does not change based off of the angle of the initial velocity, but the value of $v_0$ depends heavily upon it.
A: No, they're the same. Remember that addition is commutative and you can arrange the terms in any way (of the six possible) you wish. I've seen many physics texts start with the initial displacement, but mathematics texts will usually write a polynomial with the highest degree first.
As for the parameters, the forms take two flavors. If an object is fired vertically and you wish to model its height in time, the initial velocity is that of the vertical component. For objects fired outward at an angle, assuming no air resistance in the horizontal direction, the model
$$ y(t) = -\frac{1}{2}gt^2 + v_0 t + y_0 $$
describes the height of the object launched with initial horizontal velocity $v_0$. This model, in general, describes any projectile motion, no matter if the object is fired vertically or at an angle.
