Projectile Motion Physics Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

medium
A ball is launched horizontally at 15 m/s15 \text{ m/s} from a cliff 45 m45 \text{ m} high. How far from the base of the cliff does it land? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    Use vertical motion to find the time to fall: t=2hg=2×4510=9=3 st = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{10}} = \sqrt{9} = 3 \text{ s}
  2. 2
    Horizontal speed stays constant because there is no horizontal acceleration.
  3. 3
    Horizontal distance: x=vx⋅t=15×3=45 mx = v_x \cdot t = 15 \times 3 = 45 \text{ m}

Answer

x=45 mx = 45 \text{ m}
In projectile motion, horizontal and vertical motions are independent. The horizontal velocity remains constant (no air resistance), while the vertical motion is free fall.

About Projectile Motion

Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path.

Learn more about Projectile Motion →

More Projectile Motion Examples

Example 2 hard

A projectile is launched at [formula] at [formula] above the horizontal. What is the maximum height

Example 3 medium

A ball is kicked at [formula] at [formula]. What is the range? Use [formula].

Example 4 hard

A ball is kicked horizontally off a [formula] high cliff at [formula]. Using [formula]: (a) How long

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