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Linear Motion and Dynamics

Projectile calculations

  • 2 things to know
  • Unit AS 1

Overview

Recall questions treating the horizontal and vertical motions separately: time of flight and range for a ball leaving a bench, the components of a launch velocity at an angle, maximum height, total flight time and range over level ground, and why the horizontal component never changes.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

A football is kicked at 12 m s⁻¹ at 60° above the horizontal. Calculate the maximum height it reaches, taking g as 9.81 m s⁻².

Key term

Explain why the horizontal component of a projectile's launch velocity never appears in the calculation of its time of flight.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    A ball rolls horizontally off a bench 1.25 m high at 3.0 m s⁻¹. Calculate the time it takes to reach the floor, taking g as 9.81 m s⁻².

    Model answer

    Vertically the ball starts from rest, so 1.25 = ½ × 9.81 × t², giving t² = 0.255 and t = 0.505 s; checking back, ½ × 9.81 × 0.505² = 1.25 m, which is the height given.

  2. 2

    A ball rolls horizontally off a bench 1.25 m high at 3.0 m s⁻¹. Calculate how far horizontally from the bench it lands, taking g as 9.81 m s⁻².

    Model answer

    The time of fall is √(2 × 1.25 ÷ 9.81) = 0.505 s and the horizontal velocity is constant, so the distance is 3.0 × 0.505 = 1.5 m; checking back, 1.51 ÷ 3.0 = 0.505 s, which is the time of fall.

  3. 3

    A ball rolls horizontally off a bench 1.25 m high at 3.0 m s⁻¹. Calculate the vertical component of its velocity as it lands, taking g as 9.81 m s⁻².

    Model answer

    Using v² = u² + 2as vertically, v² = 0 + 2 × 9.81 × 1.25 = 24.5, so v = 4.95 m s⁻¹ vertically downwards; checking back, 4.95² ÷ (2 × 9.81) = 1.25 m, which is the height of the bench.

  4. 4

    A ball leaves a bench 1.25 m high with a horizontal velocity of 3.0 m s⁻¹ and lands with a downward vertical velocity of 4.95 m s⁻¹. Calculate its resultant velocity on landing.

    Model answer

    The magnitude is √(3.0² + 4.95²) = 5.8 m s⁻¹ and the direction is tan⁻¹(4.95/3.0) = 59° below the horizontal; checking back, 5.79 cos 58.8° = 3.0 m s⁻¹, which is the horizontal component given.

  5. 5

    A projectile is launched at 20 m s⁻¹ at 35° above the horizontal. Calculate the horizontal and vertical components of its initial velocity.

    Model answer

    The horizontal component is 20 cos 35° = 16.4 m s⁻¹ and the vertical component is 20 sin 35° = 11.5 m s⁻¹ upwards; checking back, √(16.4² + 11.5²) = 20.0 m s⁻¹, which is the launch speed given.

  6. 6

    A projectile is launched at 20 m s⁻¹ at 35° above the horizontal. Calculate the time it takes to reach its highest point, taking g as 9.81 m s⁻².

    Model answer

    The vertical component of launch velocity is 20 sin 35° = 11.47 m s⁻¹ and it falls to zero at the top, so t = 11.47 ÷ 9.81 = 1.17 s; checking back, 9.81 × 1.17 = 11.5 m s⁻¹, which is the initial vertical velocity.

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