Physics » One-Dimensional Motion » Vertical Projectile Motion

Equations of Motion II

Equations of Motion II

The equations of rectilinear motion that you learnt earlier can be used for vertical projectile motion, with acceleration from gravity: \(\vec{a}=\vec{g}\). Why, because the equations of rectilinear motion can be applied to any motion in a straight line with constant acceleration. If we are only considering vertical motion (up and down) it is motion in a straight line and the acceleration, \(\vec{g}\), is constant and along the same line. We use the magnitude of \(g= \text{9.8}\text{ m·s $^{-2}$}\) for our calculations.

Now would be a good time to revise both rectilinear motion and quadratic equations from Grade 10.

Remember that when you use these equations, you are dealing with vectors which have magnitude and direction. Therefore, you need to decide which direction will be the positive direction so that your vectors have the correct signs. The equations of motion that you learnt about in Grade 10 are:

\begin{align*} \vec{v}_{f}& = \vec{v}_{i}+\vec{a}t \\ \Delta \vec{x}& = \frac{\left(\vec{v}_{i}+\vec{v}_{f}\right)}{2}t \\ \Delta \vec{x}& = \vec{v}_{i}t+\frac{1}{2}\vec{a}{t}^{2} \\ \vec{v}_{f}^{2}& = \vec{v}_{i}^{2}+2\vec{a}\Delta\vec{ x } \end{align*}

In the case of vertical projectile motion, we know that the only force we will consider is gravity and therefore the acceleration will be \(\vec{a}=\vec{g}\).

\begin{align*} \vec{v}_{f}& = \vec{v}_{i}+\vec{g}t \\ \Delta \vec{x}& = \frac{\left(\vec{v}_{i}+\vec{v}_{f}\right)}{2}t \\ \Delta \vec{x}& = \vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2} \\ \vec{v}_{f}^{2}& = \vec{v}_{i}^{2}+2\vec{g}\Delta x \end{align*}\begin{align*} &\text{where}\\ \vec{v}_{i}& = \text{initial velocity} \text{(}\text{m}·{\text{s}}^{-1}\text{)} \text{at start} \\ \vec{v}_{f}& = \text{final velocity} \text{(}\text{m}·{\text{s}}^{-1}\text{)} \text{at time} t \\ \Delta \vec{x}& = \text{change in vertical position} \text{(m)} \\ t& = \text{time} \text{(s)} \\ \Delta t& = \text{time interval} \text{(s)} \\ \vec{a}&=\vec{g} = \text{acceleration due to gravity} \text{(}\text{m}·{\text{s}}^{-2}\text{)} \end{align*}

\(\Delta \vec{x}\) is displacement, the change in position. If the coordinate system you choose isn’t centred at the point from which the motion starts (i.e. you don’t have \(\vec{x}_i=0\)) then you must remember that the displacement and the final position are not the same. The position is given by \(\vec{x}_f=\vec{x}_i+\Delta \vec{x}\). If \(\vec{x}_i=0\) then \(\vec{x}_f=(0)+\Delta \vec{x}\).

Now we can apply this mathematical description of vertical motion to solving problems in the next lesson.


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