3. A boy standing on a platform throws a ball straight up with an initial velocity of 11.2ms^-1 (a) What is the maximum height above the platform reached by the ball? (b) How long does the ball take to reach maximum height? (c) What is the position of the ball after 2.0s?

(a) The maximum height reached by the ball can be calculated using the equation of motion: h = (u^2) / (2g), where u is the initial velocity and g is the acceleration due to gravity. Substituting the given values, we get h = (11.2^2) / (2*9.8) = 6.35 m.<br /><br />(b) The time taken to reach maximum height can be calculated using the equation of motion: t = u / g. Substituting the given values, we get t = 11.2 / 9.8 = 1.14 s.<br /><br />(c) The position of the ball after 2.0 s can be calculated using the equation of motion: s = ut - (1/2)gt^2. Substituting the given values, we get s = 11.2*2 - (1/2)*9.8*2^2 = 11.2 m.