Projectile Motion: Components, Max Height and Range

Projectile motion is two separate problems joined by time. Horizontal motion has constant velocity (no air resistance). Vertical motion has constant acceleration of g = 9.8 m/s² downward (use 10 on the MCAT).

Components

For launch speed v at angle θ above horizontal: v_x = v cos θ and v_y = v sin θ.

Formulas (launch and landing at the same height)

At the peak, v_y = 0 but v_x is unchanged, so the speed is not zero.

Worked example

A ball is launched at 20 m/s and 30 degrees, with g = 10. v_y = 10 m/s, v_x = 17.3 m/s. Time of flight = 2(10)/10 = 2 s. Max height = 10²/20 = 5 m. Range = 17.3 x 2 = 34.6 m.

Angles

Range is maximum at 45 degrees. Complementary angles (such as 30 and 60) give the same range. Higher angles give more height and more time.

Common traps

• Setting total velocity to zero at the peak. Only v_y is zero.
• Giving the horizontal part an acceleration. It has none.
• Using the range formula for different launch and landing heights. Solve the vertical part for time instead.

One-minute recap

• x: constant velocity. y: acceleration g.
• Peak: v_y = 0, v_x constant.
• Max range at 45 degrees; complementary angles tie.