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)