A.1 · 9c · A.1 Kinematics · projectile motion: components, s = ut + ½at², v = u + at

Exam question III: a parcel dropped from an aircraft

Split the motion into a horizontal and a vertical axis. Four parts: the path, the time, the vertical speed and the landing speed.

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Hook

An aircraft flying level releases a parcel. Does the parcel drop straight down? Does it fall for longer than one dropped from a hovering helicopter at the same height? Keep the two axes apart and the whole problem falls out.

The question · 4 parts, 7 marks

1 / 9

A parcel released from a small aircraft

The situation, not to scale. Part (i) asks you to sketch the dashed path: starts flat, curves down, lands ahead of the release point.

A small aircraft is flying horizontally at 42 m s⁻¹, at a height of 63 m above level ground. It releases a small parcel. Air resistance is negligible.

Given	Value
Velocity of the aircraft (and parcel) at release	42 m s⁻¹, horizontal
Height above the ground	63 m
Air resistance	negligible
Take g as	9.8 m s⁻²

Part	Task	Marks
(i)	Sketch the path of the parcel from the aircraft to the ground	1
(ii)	Calculate the time from release to reaching the ground	3
(iii)	Calculate the vertical component of the velocity just before it lands	1
(iv)	Determine the speed at which it lands	2

Check 1 of 6

A parcel is released from an aircraft flying level at 42 m s⁻¹. The initial VERTICAL velocity of the parcel is…

Check 2 of 6

An aircraft flying level at 55 m s⁻¹ releases a parcel 45 m above level ground. Ignoring air resistance (g = 9.8 m s⁻²), the time to reach the ground is about…

Check 3 of 6

Just before the parcel in the question lands, the horizontal component of its velocity is…

Check 4 of 6

A small aircraft flies level at 30 m s⁻¹ and releases a parcel 80 m above level ground. Ignoring air resistance, the speed at which the parcel reaches the ground is about…

Check 5 of 6

While the parcel falls, the aircraft keeps flying level at a constant 42 m s⁻¹. Ignoring air resistance on the parcel, where is the aircraft relative to the parcel?

Check 6 of 6

The time a parcel takes to reach the ground after release from a level-flying aircraft depends on…

Apply

The aircraft flies twice as fast, at 84 m s⁻¹, at the same height of 63 m. State what happens to (a) the time to reach the ground and (b) the horizontal distance the parcel travels. Explain using the two axes.

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Remember earlier lessons

From lesson A.1 · 9

Before using v² = u² + 2as with a speed in km h⁻¹, you should…

From lesson A.1 · 6

A body starts from rest with constant acceleration. If the time is doubled, the distance travelled becomes…

Summary card

Key points

• Split projectile motion into two independent axes: constant velocity horizontally, constant acceleration g vertically.
• A level release means u_y = 0. The time of fall comes from h = ½gt² and does not depend on the horizontal speed.
• The vertical component at the ground is v_y = gt (or √(2gh)). The horizontal component stays the same.
• The landing speed is the resultant √(v_x² + v_y²), found by Pythagoras. Energy gives the same answer.

Formulas

s = u_yt + ½at²
v = u + at
x = u_xt
v = √(v_x² + v_y²)
v² = u_x² + 2gh

Key terms

Component: the part of a vector along one chosen axis
Resultant: the single vector made from the components, found by Pythagoras

Common errors

• Writing u_y = 42 for a level release.
• Using the horizontal speed to find the time of fall.
• Quoting only the vertical component as the landing speed.
• Rounding t to 3.6 s before using it again.

Checks solved: 0 of 6

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