01 · A lower launch speed
A particle is launched upward at 14.7 m/s. Find time to its highest point.
Hint
Set 14.7 − 9.8t = 0.
Worked solution
t = 1.5 s.
Understand · explore · practise
Upward projections and maximum height
Find time to maximum height, distinguish rise from ground height, and calculate landing time, impact velocity and total distance.
Before you startVertical-motion signs, SUVAT and quadratic equations.
01 / Track height and signed velocity together
The particle rises while its upward velocity is positive.
Use a = −9.8 m/s² throughout free flight.
Gravity stays downward during both ascent and descent. At the highest point velocity is momentarily zero, but acceleration is still −9.8 m/s². These are not equilibrium conditions.
Rise, turn and fallExplore
A particle is projected upward at 19.6 m/s from a platform 24.5 m above ground. Take upward positive and g = 9.8 m/s²; neglect air resistance. Choose time since launch. The graph stops at impact, t = 5 s.
02 / Use zero velocity to locate the highest point
Find the time before substituting into displacement.
Launch at 19.6 m/s upward.Worked example
0 = 19.6 − 9.8t
The highest point has v = 0.
t = 2 s
This is elapsed time from launch.
s = 19.6(2) − 4.9(2²) = 19.6 m
The rise is measured above the launch point.
Watch: the apex is a turn, not a pause in gravity
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Rise without time
Find the maximum rise for that 14.7 m/s launch.
Hint
Use 0 = u² − 2gs.
Worked solution
s = 14.7²/19.6 = 11.025 m, approximately 11.0 m.
03 / Add the initial height to the vertical displacement
Rise above launch is not height above ground.
03 · Platform launch
For the model, find the maximum height above ground.
Hint
Add the platform height to the maximum rise.
Worked solution
Maximum height = 24.5 + 19.6 = 44.1 m.
04 · One second after launch
Find the model particle’s height and velocity at t = 1 s.
Hint
Use h = 24.5 + 19.6t − 4.9t² and v = 19.6 − 9.8t.
Worked solution
h = 39.2 m and v = +9.8 m/s. The particle is rising.
05 · Same height on descent
Find height and velocity at t = 3 s.
Hint
Use the same equations; gravity has not changed.
Worked solution
h = 39.2 m and v = −9.8 m/s. It is at the same height as at 1 s but moving downward.
04 / Return to the launch level before reaching lower ground
Set displacement to zero for a return to launch height.
For u = 19.6 m/s, solve s = 0.Worked example
0 = 19.6t − 4.9t² = 4.9t(4 − t)
Keep both algebraic roots initially.
t = 0 or t = 4 s
Zero is the launch; four seconds is the later return.
v(4) = −19.6 m/s
The returning speed matches the launch speed in this ideal model.
06 · Still above ground
Where is the platform-launched particle at t = 4 s?
Hint
Zero displacement refers to the launch point.
Worked solution
It is 24.5 m above ground. Returning to launch height is not ground impact.
07 · A ground-level launch
If instead launched from ground at 14.7 m/s, find its first return time to ground.
Hint
The ascent and descent to the same height take equal times.
Worked solution
Return time = 2(14.7/9.8) = 3 s, ignoring the initial t = 0 contact.
05 / Ground contact requires height zero
Use the ground height, or a negative displacement from launch.
08 · Platform landing time
For the model solve 24.5 + 19.6t − 4.9t² = 0.
Hint
Divide by 4.9 and factor.
Worked solution
t² − 4t − 5 = (t − 5)(t + 1) = 0. Reject −1 s; the first post-launch ground contact is at 5 s.
09 · Impact velocity and speed
Find the velocity and speed just before that impact.
Hint
Substitute t = 5 into v = 19.6 − 9.8t.
Worked solution
v = −29.4 m/s, so impact speed is 29.4 m/s. The downward impact speed exceeds the launch speed because the ground is below the launch point.
10 · Check without time
Check the impact speed using v² = u² + 2as.
Hint
Use s = −24.5 m and a = −9.8 m/s².
Worked solution
v² = 19.6² + 2(−9.8)(−24.5) = 864.36. Speed = 29.4 m/s; choose v = −29.4 m/s for downward impact.
06 / Split total distance at the highest point
Net displacement cannot count both parts of the journey.
11 · Distance to ground
Find total distance travelled by the model particle before impact.
Hint
Add its rise above launch to its descent from the apex to ground.
Worked solution
Distance = 19.6 + 44.1 = 63.7 m. Net displacement is −24.5 m.
12 · Two different averages
Find average velocity and average speed over the 5 s flight.
Hint
Use signed displacement for velocity and total path length for speed.
Worked solution
Average velocity = −24.5/5 = −4.9 m/s. Average speed = 63.7/5 = 12.74 m/s.
07 / State the domain and interpret the apex correctly
Free-flight equations finish at first contact.
13 · Six seconds after launch
Can the model predict the particle’s position at t = 6 s?
Hint
The particle hits the ground at 5 s.
Worked solution
Not from the free-flight model alone. A separate contact or rebound model is needed.
14 · Zero velocity at the apex
A learner says the particle has no acceleration at the top because it stops. Correct them.
Hint
Compare v and a.
Worked solution
At the apex v = 0 only instantaneously; acceleration is still −9.8 m/s². Its velocity immediately becomes negative, so it starts descending.
08 / Separate rise, height, displacement and distance
Use a single sign convention until first ground contact.
At maximum height set v = 0. Add initial height when the question asks for height above ground. Set ground height to zero for landing, then choose the signed downward impact velocity. Split the path at the apex when calculating distance.
Section 1 of 8 · Track height and signed velocity together