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Option C: The Physics of Sports

Projectiles, Bernoulli's equation and drag

  • 3 things to know
  • Component 3

Overview

Bernoulli's equation in the form used for sports aerodynamics and why a ball with sidespin curves in flight.

What you need to know

Answer each one before you reveal it.

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Key term

Explain why a shot putter releases the shot at an angle noticeably below 45°.

Key term

State Bernoulli's equation in the form used for sports aerodynamics.

Key term

Give the equation for the magnitude of the drag force on a body moving through a fluid.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why are the horizontal and vertical motions of a projectile treated separately?

    Model answer

    The only force acting once the projectile is in flight is the weight, which is vertical. The horizontal velocity is therefore unchanged while the vertical velocity changes at g, and the two motions share only the time.

  2. 2

    Give the horizontal and vertical components of the launch velocity of a projectile thrown at speed v and angle θ to the horizontal.

    Model answer

    The horizontal component is v cos θ and the vertical component is v sin θ, both in m s⁻¹. The angle is measured from the horizontal.

  3. 3

    Give the equation for the range of a projectile launched and landing at the same height.

    Model answer

    The range is R = v² sin 2θ/g, where v is the launch speed in m s⁻¹, θ the launch angle and g the gravitational field strength of 9.81 N kg⁻¹.

  4. 4

    A ball is thrown at 20 m s⁻¹ at 30° above the horizontal on level ground. Calculate its range.

    Model answer

    R = v² sin 2θ/g = 400 × sin 60°/9.81 = 400 × 0.866/9.81 = 35 m. Working back, sin 2θ = Rg/v² = 35.3 × 9.81/400 = 0.866, giving θ = 30°.

  5. 5

    For the same throw at 20 m s⁻¹ and 30°, calculate the time of flight.

    Model answer

    The time of flight is t = 2v sin θ/g = 2 × 20 × 0.500/9.81 = 2.0 s. Working back, the horizontal distance is 20 cos 30° × 2.04 = 17.3 × 2.04 = 35 m, matching the range.

  6. 6

    At what launch angle is the range of a projectile greatest on level ground, and why?

    Model answer

    The greatest range is at 45°, because sin 2θ takes its maximum value of one when 2θ = 90°. Any other angle gives a smaller value of sin 2θ and hence a shorter range.

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