A small stone is projected verically upwards from a point O with a speed of 19.6ms^-1. Modeeling the stone as a particle moving freely under gravity find the time for which the stone is more than 14.6m above O
S = 14.7, U = 19.6, V =, A = -g, T = t
using s = ut + 1/2 at^2
14.7 = 19.6t + 1/2 -g t^2
1/2 g t^2 - 19.6t + 14.7 = 0
t = (19.6 +- sqrroot(-19.62 - 4 * 0.5 * 9.8 * 14.7)) / 2 * 0.5 * 9.8
t = 1 and t = 3
Therefore the total time above 14.7 was 2 seconds
HB
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