Bodies falling freely or thrown up
Replace a by g in the equations of motion. Take g positive going down and negative going up; at the highest point v = 0.
Free fall
A body moving under the pull of gravity alone is in free fall. Near the earth, g is nearly constant. So the equations of uniformly accelerated motion still work. Just replace the acceleration a by g.
Equations of motion under gravity
| Ordinary form | Under gravity | Use it when you know |
|---|---|---|
| v = u + at | v = u + gt | u, g and t, to find v |
| s = ut + ½at² | s = ut + ½gt² | u, g and t, to find s |
| v² = u² + 2as | v² = u² + 2gs | u, g and s, to find v (no time given) |
Signs to remember
- Dropped from rest: u = 0.
- Moving down: take g = +9.8 m s⁻², because the speed increases.
- Thrown up: take g = −9.8 m s⁻², because the speed decreases.
- At the highest point, the body stops for an instant: v = 0.
- Without air resistance, the time to go up equals the time to come back down.
Worked example: a stone dropped from a bridge
A stone is dropped from a bridge and hits the water after 2 s. Find its speed as it hits the water and the height of the bridge. (g = 9.8 m s⁻²)
- Given: u = 0, t = 2 s, g = +9.8 m s⁻² (downward motion).
- v = u + gt = 0 + 9.8 × 2 = 19.6 m s⁻¹.
- s = ut + ½gt² = 0 + ½ × 9.8 × 2² = 4.9 × 4 = 19.6 m.
It hits the water at 19.6 m s⁻¹, and the bridge is 19.6 m high
Worked example: a ball thrown up
A ball is thrown straight up at 14.7 m s⁻¹. Find the greatest height it reaches and the time it takes to get there. (g = 9.8 m s⁻²)
- Given: u = 14.7 m s⁻¹, v = 0 at the top, g = −9.8 m s⁻² (upward motion).
- Height: v² = u² + 2gs gives 0 = 14.7² + 2 × (−9.8) × s.
- 14.7² = 216.09, so 19.6 s = 216.09 and s = 216.09 ÷ 19.6 ≈ 11.0 m.
- Time: v = u + gt gives 0 = 14.7 − 9.8 t, so t = 14.7 ÷ 9.8 = 1.5 s.
It rises about 11.0 m in 1.5 s, and takes another 1.5 s to fall back to the hand
The up and down journey of a ball
A coin tossed before a cricket match rises, stops for an instant and falls back. It takes as long to fall as it took to rise. On building sites, even a small bolt dropped from a high floor lands very fast, so workers wear helmets.
2 marks: “Write the equation of motion that links velocity and distance for a body under gravity. Say what u, v, g and s stand for.” v² = u² + 2gs. Here u is the initial velocity, v the final velocity, g the acceleration due to gravity and s the distance moved.
3 marks: “A ball is thrown vertically up at 19.6 m s⁻¹. Find (a) the time it takes to reach the highest point and (b) the greatest height. (g = 9.8 m s⁻²)” Method: at the top v = 0, and g = −9.8 m s⁻². (a) t = u/g = 19.6 ÷ 9.8 = 2 s. (b) s = u²/2g = 384.16 ÷ 19.6 = 19.6 m. Answer: 2 s and 19.6 m.
2 marks: “A coconut falls from the top of a tree 20 m tall. Find its speed just before it hits the ground. (g = 10 m s⁻²)” Method: u = 0, so v² = u² + 2gs = 0 + 2 × 10 × 20 = 400. So v = √400 = 20 m s⁻¹.
Up is minus, down is plus, and the top is where v is zero.