Analyze projectile motion by treating the horizontal and vertical motions independently: constant horizontal velocity and vertical free fall, linked only by the common time of flight.

What this topic is asking

Projectile motion extends kinematics to two dimensions, and the Regents tests one central idea: the horizontal and vertical motions are independent and can be solved separately, joined only by the shared time of flight. The Physical Setting/Physics course focuses on the cleanest case, a projectile launched horizontally, and asks you to find the time in the air, the horizontal range, and the landing speed. Understanding the independence is worth more marks than memorizing any formula, because the formulas are just the one-dimensional kinematic equations applied to each axis.

The independence of horizontal and vertical motion

A classic demonstration of this independence: a ball thrown horizontally and a ball dropped from the same height at the same instant land at the same time, because their vertical motions are identical. The thrown ball travels sideways as well, but its fall is unaffected by its horizontal speed.

Horizontal motion: constant velocity

Since no horizontal force acts, the horizontal velocity $v_x$ never changes. The horizontal distance is therefore simply

This is the constant-velocity case, not an accelerated one, so you do not use $\tfrac{1}{2}at^2$ horizontally. For a horizontally launched projectile, $v_x$ is just the launch speed.

Vertical motion: free fall

The vertical motion obeys the free-fall kinematics, with acceleration $g$ downward. For a projectile launched horizontally, the initial vertical velocity is zero, so

The vertical drop $d_y$ determines the time of flight, the bridge between the two motions.

Finding range and landing speed

The order is fixed: vertical first (to get the time), then horizontal (to get the range). Trying to find the time from the horizontal data fails, because the horizontal motion alone never "runs out", it is the ground that stops the projectile, and that is a vertical condition.

Reference Tables note

The Reference Tables print the one-dimensional kinematic equations ($v_f = v_i + at$, $d = v_i t + \tfrac{1}{2}at^2$, $v_f^2 = v_i^2 + 2ad$) and the constant $g$, which are all you need; there is no special projectile-motion formula on the tables. You supply the strategy of splitting the motion into perpendicular components, and you use the vector-combination tools from vectors and scalars to find the landing speed.

Try this

Q1. State why the horizontal velocity of a projectile stays constant (neglecting air resistance). [1 point]

• Cue. There is no horizontal force, so by Newton's first law the horizontal velocity does not change.

Q2. A ball launched horizontally takes $2.0$ s to land and has a horizontal velocity of $8.0$ m/s. Calculate the horizontal distance travelled. [2 points]

• Cue. $d_x = v_x t = (8.0)(2.0) = 16$ m.