A ball is thrown up with an initial velocity of 8 m/s and initial height of 1.5m above the ground. Calculate the maximum height the ball reaches and the time it takes to get there.
Travelling upwards
u = 8m/s v = 0 m/s a = -9.8m/s S = ? t = ?
v = u + at v2 = u2 + 2a(S - 1.5)
0 = 8 -9.8t 0 = 64 + 2*(-9.8)*(S - 1.5)
9.8t = 8 19.6*(S - 1.5) = 64
t = 0.816s (3.sf) S - 1.5 = 3.265m
S = 4.765m (3.sf)
ES
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