A baseball is thrown vertically upward with an initial velocity of 14 mps. Find the: a) velocity with which it strikes the ground b) maximum height it can reach c) time taken to reach that height

Hello!
The movement of a body thrown vertically upwards is uniformly accelerated (or decelerated, if you wish). The speed V of a body is equal to V(t) = V_0 - gt, where t is a time, V_0 is an initial upward speed and g approx 9.8 m/s^2 is the gravity acceleration. The minus sign before g reflects the fact that gravity acceleration is directed downwards.
The height of a body is H(t) = H_0 + V_0 t - (g t^2)/2, where H_0 is the initial height (zero in our problem). A baseball strikes the ground at t_1gt0 such that H(t_1) = 0. It is obvious that t_1 = (2 V_0)/g. The velocity at this moment is V(t_1) = V_0 - g*(2 V_0)/(g) = -V_0, i.e. the same but downwards.
The maximum height is reached at the parabola vertex, and the time is t_2 = -b/(2a) = V_0/(g) approx 1.43 (s) (the half of t_1, actually). The height at that time is (V_0)^2/(2g) approx 10 (m).
The answers: a) 14 m/s downwards, b) 10 m, c) 1.43 s.