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Projectile Motion overview

Topic 3 of 4

Launch and land at different heights

Set the vertical equation equal to the actual landing height. Returning to the launch height and reaching lower ground are different events.

Assume ideal flight: uniform downward g = 10 m/s2, negligible air resistance and no force after launch except weight. Choose right and up as positive, with t = 0 at launch and ground level at y = 0.

The initial position is (0, 20 m), with ux = 12 m/s and uy = 9.0 m/s. These components give launch speed √(122 + 92) = 15 m/s at tan-1(9/12) = 36.9 degrees above horizontal.

As in the horizontal-launch example, horizontal velocity is constant and vertical acceleration is -g. With t in s, the numerical equations give positions in m and velocities in m/s:

x = 12t
y = 20 + 9t - 5t2
vx = 12
vy = 9 - 10t

Returning to the launch height is not landing

Here x = 12t and y = 20 + 9t - 5t2, with metres and seconds. Up is positive. Filled dots show the supplied model every 0.50 s; the special events are marked separately.

A projectile launched twenty metres above the groundThe horizontal and vertical position axes use the same scale, 7.3 drawing units per metre. A continuous parabola starts at zero metres horizontal position and twenty metres height. Filled dots at times zero, 0.5, 1.0, 1.5, 2.0, 2.5 and 3.0 seconds follow the specified trajectory. A hollow diamond marks the exact top at time 0.90 seconds, position 10.8 metres and height 24.05 metres; it is distinct from the nearby one-second dot. A hollow circle marks the descending crossing of the dashed twenty-metre launch-height line at time 1.80 seconds and horizontal position 21.6 metres. The final endpoint lies on the ground at the exact future landing time, approximately 3.093171 seconds and range 37.118055 metres, after the three-second dot. In a separate inset for the top, a blue velocity arrow points right and a purple weight arrow points down. The velocity there is twelve metres per second, so the object has not stopped. The inset arrows have no common scale with positions or with each other.0102030400510152025Height y / mHorizontal position x / mAt the topvW

Diamond: exact top. t = 0.90 s; (x, y) = (10.8, 24.05) m. Velocity is 12 m/s right; weight is still downward.

Hollow circle: launch height again. t = 1.80 s; (x, y) = (21.6, 20.0) m. The ground is still below.

Ground endpoint: landing. t = 3.093171... s and x = 37.118055... m. The last regular dot is at 3.0 s, before landing.

Both position axes use the same metre scale. The small inset distinguishes velocity from weight; its arrow lengths are schematic.

The marked positions at half-second intervals lie on the supplied model trajectory. The exact top at 0.90 s is distinct from the 1.0 s point, which is already descending. Ground and launch height mark different levels.

1. Choose the future landing root

The ground is y = 0, so:

0 = 20 + 9t - 5t2
5t2 - 9t - 20 = 0
t = (9 ± √481)/10 s

The roots are approximately +3.093 s and -1.293 s. Only the positive root describes the landing after the stated launch. The negative root belongs to a mathematical extension of the equation to an earlier time; it is not this flight's landing time.

The horizontal range is 12 × 3.093171... = 37.12 m as a working value. Keep the unrounded time while calculating later quantities.

2. Combine the landing velocity components

vx = +12 m/s
vy = 9 - 10[(9 + √481)/10]
= -√481 m/s ≈ -21.93 m/s

The negative vertical component means downward motion. Speed and direction follow from both components:

v = √(122 + 481) = 25.0 m/s
α = tan-1(√481/12) ≈ 61.3°

The landing direction is 61.3 degrees below horizontal, down and right. The result 21.93 m/s is the magnitude of the vertical component, not the full speed.

3. Distinguish the top and return to launch height

At the top, vy = 0, so 9 - 10t = 0 gives t = 0.90 s. Its position is x = 12(0.90) = 10.8 m and y = 20 + 9(0.90) - 5(0.90)2 = 24.05 m. Speed is still 12 m/s because the horizontal component remains.

On descent, the projectile returns to y = 20 m when 9t - 5t2 = 0 gives the nonzero time 1.80 s. It is then at x = 21.6 m, still above the ground. The expression 2uy/g gives this same-height return, not the time to the lower landing level.

Two velocity components share one elapsed time

These graphs describe the launch from 20 m with ux = 12 m/s and uy = 9.0 m/s. Up and right are positive. Both traces end at the same landing event.

Horizontal velocity stays at +12 m/s

Horizontal velocity stays at +12 m/sHorizontal velocity is twelve metres per second throughout the flight. The horizontal time axis lies at velocity zero. The blue line starts at time zero and stops at the exact landing time, approximately 3.093171 seconds; it does not continue after impact. The graph shares its elapsed-time scale and landing endpoint with the vertical-velocity graph.0123.09305101215vx / m/sElapsed time t / s

Horizontal acceleration is zero throughout ideal flight. Constant vx does not make the total velocity vector constant.

Vertical velocity changes at -10 m/s each second

Vertical velocity changes at -10 m/s each secondUp is positive. Vertical velocity starts at positive nine metres per second, falls linearly with gradient negative ten metres per second squared, crosses zero at 0.90 seconds and reaches negative square root of 481, approximately negative 21.931712 metres per second, at landing. The arrowed horizontal time axis is at velocity zero, so negative velocities appear below it. Bottom time labels refer to the same grid. The line stops at the exact landing time, approximately 3.093171 seconds, matching the horizontal-velocity graph.0123.093-25-20-1009vy / m/sElapsed time t / svy = 0 at 0.90 s

The line crosses the zero-velocity time axis at the top. Its gradient remains -10 m/s2 there. At landing, vy = -21.931712... m/s.

The dotted landing guide is at t = (9 + √481)/10 s = 3.093171... s. The short axis label is rounded; the plotted endpoint uses the full value. Neither line describes the collision after landing.

These are velocity-time graphs for the same flight. The horizontal component stays at +12 m/s; the vertical component has gradient -10 m/s2 and crosses zero at 0.90 s. Both end at the actual landing time.

4. Check the speed using energy

Let the ball's mass be 0.20 kg. With negligible drag, the ball-Earth system's decrease in gravitational potential energy becomes an increase in kinetic energy during flight.

Ek,launch = ½(0.20)(15)2 = 22.5 J
ΔEp = (0.20)(10)(0 - 20) = -40.0 J
Ek,landing = 22.5 + 40.0 = 62.5 J

Then v = √(2Ek/m) = √(2 × 62.5/0.20) = 25.0 m/s, agreeing with the component calculation. Energy finds the speed here; it has not by itself found the flight time or direction.

The extra digits in these model calculations help compare methods and locate graph features. To two significant figures, the landing results are 3.1 s, 37 m, 25 m/s and 61 degrees below horizontal. Retain the unrounded time for intermediate calculations, and report final results to precision justified by the supplied quantities.

Optional check After launch at t = 0, a projectile has y = 20 + 9t - 5t^2, with y in metres and t in seconds. The ground is y = 0. Which time describes landing after launch?
After launch at t = 0, a projectile has y = 20 + 9t - 5t^2, with y in metres and t in seconds. The ground is y = 0. Which time describes landing after launch?

Read the equal-time position data

Calculated model values, not camera observations
t / sx / my / mvy / m s-1
0.0020.00+9
0.5623.25+4
1.01224.00-1
1.51822.25-6
2.02418.00-11
2.53011.25-16
3.0362.00-21

Every 0.50 s adds 6.0 m horizontally, giving Δx/Δt = 12 m/s. Vertical changes are different: from 0 to 0.50 s, average vertical velocity is (23.25 - 20.00)/0.50 = 6.5 m/s, associated with midpoint time 0.25 s. From 0.50 to 1.00 s it is (24.00 - 23.25)/0.50 = 1.5 m/s, at midpoint time 0.75 s.

The velocity change divided by the midpoint-time difference is (1.5 - 6.5)/(0.75 - 0.25) = -10 m/s2. This agrees exactly with the uniform-acceleration model used to calculate the table. Real video readings have uncertainty and must be assessed using the calibration and timing method; a generated table is not experimental verification.