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 14 metres per second satisfies 0 equals 196 minus 19.6 s, giving a greatest height of 10 metres.
A dropped object starts with u equals 0, so the distance fallen is half g t squared. Falling 19.6 metres takes 2 seconds, since 19.6 equals 4.9 t squared. Falling 44.1 metres takes 3 seconds, since 44.1 equals 4.9 times 9.
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 4 or negative 1: the 4 seconds is the landing, and the negative 1 belongs to the parabola 1 second before the throw, so you reject it.
Keep one sign convention for the whole solution and never flip halfway. A ball thrown up at 15 metres per second climbs for about 1.5 seconds, reaches about 11.5 metres, and returns to the hand after about 3.1 seconds.
Gravity pulls at 9.8 down, the top means v is zero, and a negative root belongs before the clock.
2 · Watch
Take it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.