Projectile Motion Physics Example 3

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Example 3

medium
A ball is kicked at 20 m/s20 \text{ m/s} at 45°45°. What is the range? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    Use the range formula: R=v02sin⁡2θg=400sin⁡90°10=40010=40 mR = \frac{v_0^2 \sin 2\theta}{g} = \frac{400 \sin 90°}{10} = \frac{400}{10} = 40 \text{ m}
  2. 2
    At 45°45°, sin⁡90°=1\sin 90° = 1, which gives the maximum possible range for a given launch speed.

Answer

R=40 mR = 40 \text{ m}
The range formula R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g} shows that 45°45° gives the maximum range for level ground launches, since sin⁡2θ\sin 2\theta is maximized at θ=45°\theta = 45°.

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 2 hard

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

Example 4 hard

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

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