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Vertical projectile motion

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3.3 Chapter summary (ESCK2)

Projectiles are objects that move through the air. In vertical projectile motion we deal with objects that fall under the influence of gravity and only vertically.

Objects that move up and down (vertical projectiles) on the Earth accelerate with a constant acceleration \(\vec{g}\) which is approximately equal to \(\text{9,8}\) \(\text{m·s $^{2}$}\)directed downwards towards the centre of the earth.
 The time it takes an object to rise to its maximum height, is the same as the time it will take to fall back to its initial height. The magnitude of the velocity will also be the same but the direction will be reversed. This is known as time symmetry and is a consequence of uniform gravitational acceleration.

The equations of motion can be used to solve vertical projectile problems.
\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 \vec{x} \end{align*} 
Graphs for vertical projectile motion are similar to graphs for motion at constant acceleration.
Physical Quantities  
Quantity  Vector  Unit name  Unit symbol 
Position (\(x\))    metre  m 
Displacement (\(\Delta \vec{x}\))  \(\checkmark\)  metre  m 
Instantaneous velocity (\(\vec{v}\))  \(\checkmark\)  metre per second  \(\text{m·s$^{1}$}\) 
Instantaneous speed (\(v\))    metre per second  \(\text{m·s$^{1}$}\) 
Instantaneous acceleration (\(\vec{a}\))  \(\checkmark\)  metre per second per second  \(\text{m·s$^{2}$}\) 
Magnitude of acceleration (\(a\))    metre per second per second  \(\text{m·s$^{2}$}\) 
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