Vertical motion under gravity
1) The acceleration always acts downwards whatever direction the particle is moving.
2) We assume that there is no air resistance, that the object is not spinning or turning and that
the object can be treated as a particle.
3) We assume that the gravitational acceleration remains constant and is 9.8 m s-2.
4) Always state which direction (up or down) you are taking as positive.
Example: A ball is thrown vertically upwards from O with a speed of 14 m s-1.
(a) Find the greatest height reached.
(b) Find the total time before the ball returns to O.
(c) Find the velocity after 2 seconds.
Solution: Take upwards as the positive direction. ↑+
(a) At the greatest height, h, the velocity will be 0 and so we have
↑+ u = 14, v = 0, a = –9.8 and s = h (the greatest height).
Using v2 = u2 + 2as we have 02 = 142 + 2 x (–9.8) x h
→ h = 196 ÷ 19.6 = 10.
Answer: Greatest height is 10 m.
(b) When the particle returns to O the displacement, s, from O is 0 so we have
↑+ s = 0, a = –9.8, u = 14 and t = ?
Using s = ut + 1/2at2 we have 0 = 14t – 1/2 x 9.8t2
→ t(14 – 4.9t) = 0
→ t = 0 (at start) or t = 2(6/7) seconds.
Answer: The ball takes 2(6/7) seconds to return to O.
Using v = u + at we have
↑+ v = 14 – 9.8 x 2
→ v = –5.6.
Answer: After 2 seconds the ball is travelling at 5.6 m s-1
downwards.