2.5.1 Projectile Motion

When we throw an object non-vertically (like when we shoot a basketball) into the air, the object will take on a curved path. Because gravity. Gravity is the reason projectile motion is not rectilinear motion. Does this mean we can’t use the equations of motion? Does this mean we have to draw vector triangles to solve projectile motion?

Fortunately, some smart people have noticed that since gravity acts vertically downward, it only affects the vertical motion, and leaves the horizontal motion untouched. (We are assuming that the effect of air resistance is negligible, of course.)

So, the smart thing to do, is for us to mentally break a projectile motion into two rectilinear motions: the vertical one, which is a uniform acceleration motion (with constant downward acceleration of g=9.81\text{ m }{{\text{s}}^{{-2}}}), and a horizontal motion which is constant speed motion. In fact, we can visualize a projectile motion as a vertical throw that cruises horizontally (at a constant speed).

To illustrate further, the diagram below shows how the velocity varies during a projectile motion. The tangents to the curve (drawn in green) correspond to the velocity of the projectile motion at each point. Notice that while the vertical component (drawn in red) is changing (in the downward direction) at a constant rate, the horizontal component (in blue) is constant!

This is why the easiest way to calculate a projectile motion is to calculate its vertical and horizontal motion separately. For a projectile launched at speed u at an angle of θ, we always resolve u to obtain the initial horizontal and vertical components of velocity, ux and uy.

\begin{array}{l}{{u}_{x}}=u\cos \theta \\{{u}_{y}}=u\sin \theta \end{array}

Horizontally, the projectile moves forward at a constant speed. The horizontal component of velocity remains as the initial horizontal velocity throughout the motion. So there is no need to further calculate any horizontal velocity. As for the horizontal displacement, the calculation is as simple as \text{distance}=\text{speed}\times \text{time}.

{{v}_{x}}={{u}_{x}}

{{s}_{x}}={{u}_{x}}t

Vertically, the projectile is moving under the influence of (constant) gravity. So the vertical component of velocity is changing at a constant rate of g=9.81\text{ m }{{\text{s}}^{{-2}}}downward. So the vertical motion must be done using the equations of motion

{{v}_{y}}={{u}_{y}}-gt

{{s}_{y}}={{u}_{y}}t-\frac{1}{2}g{{t}^{2}}

{{v}_{y}}^{2}={{u}_{y}}^{2}-2g{{s}_{y}}

Video Explanation 

Divide and Conquer

Demonstration 

Man on Travelator 

Ting Ting Ting

Coins Swept Off Table

Interesting

Shoot the Monkey

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