Free fall and projectiles

A projectile runs two independent one-dimensional problems at once, and gravity only writes in one of them.

Horizontal and vertical split

Gravity acts only vertically, so the horizontal motion is constant velocity and the vertical motion is free fall with a = -9.8 m/s^2. The two directions share exactly one quantity: the time of flight. A ball dropped and a ball thrown horizontally from the same height hit the ground together.

x = v_{0x} t, \quad y = v_{0y} t - \tfrac{1}{2}gt^2

Range and height

For a launch at speed v0 and angle theta over level ground, the vertical and horizontal components are v0y = v0 sin(theta) and v0x = v0 cos(theta). The time of flight is 2 v0y/g, so the range is v0x times that. Maximum height uses v^2 = v0^2 + 2ax with final vertical velocity zero.

R = \frac{2 v_0^2 \sin\theta \cos\theta}{g}, \quad h_{\max} = \frac{v_0^2 \sin^2\theta}{2g}

Worked example

A ball is kicked at 20 m/s at 30 degrees above horizontal. The components are v0x = 20 cos(30) = 17.3 m/s and v0y = 20 sin(30) = 10 m/s. Time of flight is 2(10)/9.8 = 2.04 s, so the range is 17.3 times 2.04 = 35.4 m. Max height is v0y^2/(2g) = 100/19.6 = 5.1 m.

R = 17.3 \times 2.04 = 35.4 \text{ m}, \quad h = \frac{10^2}{19.6} = 5.1 \text{ m}

Flashcard

v_{0x} = v_0\cos\theta, \ v_{0y} = v_0\sin\theta

Splitting a launch velocity into components

Every projectile starts by decomposing the initial speed into horizontal and vertical pieces. The horizontal piece stays constant through the whole flight; the vertical piece is what gravity steadily erodes.

Check your understanding

A dart is thrown horizontally at 15 m/s from a 20 m high cliff. How far from the base of the cliff does it land?

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