# Confusion regarding 2 dimension motion

Projectile motion consist of horizontal and vertical motion.
The horizontal motion consists only of constant velocity, that is, the velocity of the object
The vertical motion consists only of constant acceleration, that is, acceleration due to gravity.

Yet....

we have vx = vcos
and
vy = vsinΘ

can someone the concept?

SteamKing
Staff Emeritus
Homework Helper
Yes, your statement "The vertical motion consists only of constant acceleration" is not accurate.

If a projectile is fired at an angle to the horizon, the initial vertical velocity Vy = V * sin (theta)

During flight, the Vy component of projectile velocity decreases under the action of gravity until Vy = 0, at which point the projectile begins to fall back to earth.

So vertical motion has vertical velocity expressed mathematically as vsinΘ+0.5gt^2
whereas
horizontal motion has velocity expressed as vcosΘ with no acceleration due to gravity?

PeroK
Homework Helper
Gold Member
2020 Award
The angle θ relates to the initial velocity. After that the projectile will be moving at a different angle.

So, in these problems, θ is the initial angle that determines the initial vertical and horizontal velocity.

If you want to, you could look at the angle θ(t) as this changes with time as the projectile moves. This would depend on the initial θ = θ(0), the initial angle of the projectile.

The important point is that θ is the initial angle.

Regardless of the change of θ, the equation vsinΘ+0.5gt^2 is correct expression for vertical component of motion. Here θ is initial angle.

SteamKing
Staff Emeritus
Homework Helper
Gravity acts toward the center of mass of the attracting body. This is why the vertical velocity is affected by gravity, but not the horizontal velocity.

jtbell
Mentor
It's important to distinguish the initial speed and direction (angle) at t = 0 from the speed and direction at a later time t. We often do this by writing the initial values as v0 and θ0.

At time t the velocity components are

$$v_x = v_0 \cos \theta_0 \\ v_y = v_0 \sin \theta_0 - \frac{1}{2}gt^2$$

whereas the speed and direction are

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0 \cos \theta_0)^2 + \left(v_0 \sin \theta_0 - \frac{1}{2}gt^2\right)^2}\\ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left( \frac {v_0 \sin \theta_0 - \frac{1}{2}gt^2} {v_0 \cos \theta_0}\right)$$