Both projectiles are acted on by the same forces: air resistance, gravity, wind, etc. (depending on the situation). But for the purpose of standard projectile motion problems, the main difference is that the vertical projectile will have *no x-component of motion.*

The time-in-air of both projectiles will be the same,...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Both projectiles are acted on by the same forces: air resistance, gravity, wind, etc. (depending on the situation). But for the purpose of standard projectile motion problems, the main difference is that the vertical projectile will have *no x-component of motion.*

The time-in-air of both projectiles will be the same, but the projectile fired vertically will land in the same spot that it was fired from. For the projectile fired at an angle, you'll have to compute the x- and y- components separately.

`x=x_0+v_xt` assuming a constant horizontal velocity

`v_x=v_0cos\theta` where `v_0` is the initial velocity (at angle of projection)

`v_{oy}=v_0sin\theta` and `v_y=v_{oy} - g t`

`y= y_o + v_{0y} t - 1/2 g t^2`

These same equations can be used for both projectiles, but the x-component will zero-out for the vertically fired projectile `(cos90^o=0)`

Note that gravity is only applicable in the vertical direction (and acts in the -ve vertical direction).

See Image, credit: Western Washington University