Vertical Motion Under Gravity
Apply the constant-acceleration equations to an object thrown or dropped, taking the acceleration as g downwards and keeping one sign convention for the whole solution.
What a learner can do afterwards
- Find the greatest height of a ball thrown upwards, using velocity zero at the top
- Find the time for a dropped object to fall a given height
- Explain what a negative root means when an object lands below its starting point
1 · Read
Falling is constant acceleration wearing a familiar number: 9.8 metres per second squared downwards. Take upwards as positive and gravity enters as negative 9.8, which keeps every sign honest. At the top of a throw the velocity is exactly zero, so a ball thrown up at 21 metres per second satisfies 0 equals 441 minus 19.6 s, giving a greatest height of 22.5 metres.
A dropped object starts with u equals 0, so the distance fallen is half g t squared. Falling 4.9 metres takes 1 second, since 4.9 equals 4.9 t squared. Falling 78.4 metres takes 4 seconds, since 78.4 equals 4.9 times 16.
When the landing sits below the start, the equation is quadratic and one root is negative. A ball thrown from a cliff lands when t is 5 or negative 2: the 5 seconds is the landing, and the negative 2 belongs to the parabola 2 seconds before the throw, so you reject it.
Keep one sign convention for the whole solution and never flip halfway. With no air resistance the fall back to the hand takes as long as the climb, so the whole flight is twice the climb time. A ball thrown up at 12 metres per second climbs for about 1.2 seconds, reaches about 7.3 metres, and is back in the hand after about 2.4 seconds.
Gravity pulls at 9.8 down, the top means v is zero, and a negative root belongs before the clock.
The rewritten readings were written for this dataset. Practice items adapted from WeBWorK Open Problem Library problems by Brandon Lostracco (Brock University) (CC BY-NC-SA 3.0).
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