Projectile Motion Physics Example 2

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

Example 2

hard
A projectile is launched at 40 m/s40 \text{ m/s} at 30°30° above the horizontal. What is the maximum height and horizontal range? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    Components: vx=40cos⁡30°=34.64 m/sv_x = 40\cos 30° = 34.64 \text{ m/s}, vy=40sin⁡30°=20 m/sv_y = 40\sin 30° = 20 \text{ m/s}.
  2. 2
    Maximum height: h=vy22g=40020=20 mh = \frac{v_y^2}{2g} = \frac{400}{20} = 20 \text{ m}
  3. 3
    Total flight time: t=2vyg=4010=4 st = \frac{2v_y}{g} = \frac{40}{10} = 4 \text{ s}.
  4. 4
    Range: R=vx⋅t=34.64×4=138.56 mR = v_x \cdot t = 34.64 \times 4 = 138.56 \text{ m}

Answer

hmax⁡=20 m,R≈138.6 mh_{\max} = 20 \text{ m}, \quad R \approx 138.6 \text{ m}
Projectile motion at an angle requires resolving the initial velocity into components. The vertical component determines height and flight time; the horizontal component determines range.

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 1 medium

A ball is launched horizontally at [formula] from a cliff [formula] high. How far from the base of t

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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