Projectile Motion Problems -- Type 2
A “Type 2” Projectile Motion problem is one where a projectile is launched with a 2D initial
velocity vector, and it returns back to the same height when it lands. In a general sense, we will
visualize the following picture:
In this diagram, the initial velocity v0 is directed at an angle 𝜃, and it must be decomposed into
its x- and y-components, as we have done in the past for any vector:
Additionally, notice that the vertical displacement ∆y = 0. We should recall from Free-fall that
this implies a few things:
1. When the object lands, it has a final vertical velocity that is equal in Magnitude, but
opposite in direction, to the initial vertical velocity. In other words,
2. The time from bottom to top is equal to the time from top to bottom; these are both equal
to 1/2 the total time in the air.
** Special Equation! ** In the case of a Type 2 problem, we can always determine the total
horizontal displacement, called the “Horizontal Range”, using the following equation:
Two important things to note about this equation:
1. The “v0” in the equation is the entire magnitude of the initial velocity; is is NOT the x- or
the y-component of the initial velocity.
2. The “Sin2𝜃” in the equation literally means we double the angle; if the angle is given as
30 degrees, we will use Sin60.
An interesting side effect of this Horizontal Range equation is the fact that, when launched with
the same initial velocity, projectiles will have the same Horizontal Range whenever the launch
angles are Complementary (they add up to 90 degrees).
One additional note: when a projectile reaches its maximum height, its velocity is NOT zero;
this is because there is a horizontal velocity that does not change. The vertical velocity is zero at
the maximum height, but the horizontal velocity will still be equal to v0x.