Projectile problem solved using SUVAT--- a) Maximum Height - consider only vertical components: u = 4 m/s, a = -9.81 m/s2, v = 0 ( at maximum height vertical speed is 0), s = ? - Use v2 = u2 + 2a (s-1) = 0 = 42 + 2 ( -9.81) (s-1) = 16 - 19.625(s-1). s = 16 / 19.625 +1 = 1.815m --- b) Velocity hitting the ground - consider both components: horizontal component of velocity is constant as no acceleration. Vertical component use v2 = u2 + 2as with u = 4m/s, a = -9.81m/s2 and s = -1m. v = - sqrt ( 16 + 19.62 ) = - 5.968m/s ( as pointing down), Velocity = (7i - 5.968j)m/s , speed = 9.199m/s. --- c) Distance travelled requires time - found from vertical component and v = u + at, t = - 5.968 - 4 / -9.81 = 1.016s, substituting into the horizontal component using s = ut, s = 7.113m.