01 · Higher drop
Find fall time and impact speed from44.1 m, starting from rest.
Hint
Use H = 4.9t².
Worked solution
t² = 9, so t = 3 s. Impact speed = 29.4 m/s.
Understand · explore · practise
Drops and downward projections
Calculate falling times and impact speeds for drops and downward throws, with consistent signs and a model that ends at contact.
Before you startVertical-motion signs, SUVAT and quadratic equations.
01 / Choose initial velocity from the wording
Released from rest means u = 0; thrown downward generally does not.
Use downward positive for these calculations.
Let H be the positive starting height above ground. At impact the downward displacement is H, a = +9.8 m/s² and initial downward velocity u ≥ 0. Neglect air resistance and use the equations only up to first ground contact.
Drop or throw downwardExplore
Take downward as positive, g = 9.8 m/s² and negligible air resistance. Choose a starting height and initial downward speed. The height graph ends at impact.
02 / A drop has a simple time equation
Set the displacement equal to the height fallen.
A particle is dropped from19.6 m.Worked example
19.6 = ½ × 9.8 × t²
Its initial velocity is zero.
t² = 4, so t = 2 s
Reject the negative time.
v = 9.8 × 2 = 19.6 m/s downward
This is the velocity just before impact.
02 · Two seconds into that drop
How high is the particle after2 s in question1?
Hint
Subtract the distance fallen from44.1.
Worked solution
It has fallen19.6 m, leaving height24.5 m. It is still in free flight.
03 / You can find impact speed without first finding time
The squared equation omits time.
Watch: height decreases and downward speed increases
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Speed from height
Find the impact speed from19.6 m directly.
Hint
Use v² = 2gH.
Worked solution
v² = 2 × 9.8 × 19.6 = 384.16, so speed = 19.6 m/s.
04 · Recover the height
A particle released from rest reaches the ground at24.5 m/s. Find its release height.
Hint
Rearrange H = v²/(2g).
Worked solution
H = 24.5²/19.6 = 30.625 m, approximately30.6 m.
04 / A downward throw adds an initial-speed term
Use H = ut + ½gt².
A particle is thrown down at4.9 m/s from29.4 m.Worked example
29.4 = 4.9t + 4.9t²
The initial-speed term is not zero.
t² + t − 6 = (t + 3)(t − 2) = 0
Divide by4.9 and solve.
t = 2 s; v = 4.9 + 9.8 × 2 = 24.5 m/s
Reject t = −3 s.
05 · Another downward throw
Find time and impact speed for u = 9.8 m/s from39.2 m.
Hint
Divide the displacement equation by4.9.
Worked solution
t² + 2t − 8 = (t + 4)(t − 2) = 0. The physical time is2 s and impact speed is29.4 m/s.
06 · Compare at the same height
Would a drop from29.4 m take longer than the4.9 m/s downward throw?
Hint
For the drop, t² = 6.
Worked solution
Yes. The drop takes√6 s, about2.45 s, compared with2 s for the downward throw.
05 / Another axis must give the same physical answer
Change all signed quantities consistently.
07 · Upward-positive version
For the4.9 m/s downward throw from29.4 m, state u, a and impact displacement with upward positive.
Hint
Downward quantities are negative.
Worked solution
u = −4.9 m/s, a = −9.8 m/s² and s = −29.4 m.
08 · Signed impact velocity
Using that axis, find v at the2 s impact time.
Hint
v = u + at.
Worked solution
v = −4.9 − 9.8 × 2 = −24.5 m/s. Its speed is24.5 m/s; both axes describe the same downward motion.
06 / Subtract cumulative distances for a later interval
Distance fallen in each second increases.
09 · Final second of a drop
For the drop from44.1 m, find the distance fallen during the final second before impact.
Hint
Impact is at3 s; compare distances at3 and2.
Worked solution
Distance = 44.1 − 19.6 = 24.5 m. This is not one third of the total height.
10 · Average speed
Find average speed over the whole3 s drop from44.1 m.
Hint
Divide total distance by total time.
Worked solution
Average speed = 44.1/3 = 14.7 m/s, half the final speed because velocity increases linearly from zero.
11 · First half-second
How far does a dropped particle fall in0.5 s?
Hint
Square the time before multiplying.
Worked solution
Distance = 4.9 × 0.5² = 1.225 m, approximately1.23 m.
07 / Stop at impact and retain the model assumptions
A continuation below ground is not a physical trajectory.
12 · Beyond impact
Can the formula for a19.6 m drop be used to report the particle’s height at3 s after release?
Hint
It hits the ground at2 s.
Worked solution
Not without a new post-impact model. Substitution would produce a negative height, but that is only an invalid extrapolation of the free-flight equation.
13 · Negative root
Why is the t = −3 s root rejected for the downward-throw example?
Hint
The model begins at release, t = 0.
Worked solution
It describes an extrapolated earlier event outside the stated domain. The required flight duration must be nonnegative.
14 · Real-object discrepancy
A light object falls more slowly than this prediction. Name an assumption to reconsider.
Hint
Which force was ignored?
Worked solution
Air resistance may be significant. The constant free-fall acceleration model then needs revision; do not simply change signs to force agreement.
08 / Use the wording, axis and contact condition
Distinguish impact speed from signed impact velocity.
For a drop set u = 0; for a downward throw retain u. Set displacement equal to the height fallen using a consistent axis. Solve for nonnegative time, calculate the pre-impact speed and stop the flight model at ground contact.
Section 1 of 8 · Choose initial velocity from the wording