Question

Explain how $u\cos \theta$ is the horizontal component and $u\sin \theta$ the vertical component when a projectile is projected at an angle $\theta$ from the horizontal.

Answer 1

Let the x-axis be along horizontal and y-axis along vertical as hown in figure below.

From the right triangle,
$\sin \theta =\frac{u_y}{u}$
$\Rightarrow u_y=u\sin \theta$

Similarly, $\cos \theta =\frac{u_x}{u}$
$\Rightarrow u_x=u\cos \theta$

The net initial velocity can be written as
$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$