ok, so I am trying to learn these ones too. here is what i know. 14m/s vertical vector component of velocity. how high will the ball travel? well, this a kinematic equation problem.
Vo = initial velocity = 14m/s. (says in the question)
V = final velocity (when the ball is at the top of its flight path, this is) = 0
a = acceleration = -9.8m/s^-2 (only force acting on this object)
t = time = ?
x = change in x, or change in displacement, or change in distance vertically = ?
but with three knowns, we can find a fourth. in this case it would be good to see how high the ball is going to go, so we want to find 'change in x'
we know that
V^2 = Vo^2 + 2*a*x
this is on of the five kinematic equations derived from the ideas that velocity = change in distance/ change in time, and, acceleration = change in velocity/ change in time
so, rearanging mathematically this formula, to isolate x, we get
((V^2) - (Vo^2)) /2*a = x
now we can substitiute in the values we know to solve for x
((0-14^2))/2*-9.8 = x
10 = x is what i get. so this tells us the ball will get to a maximum height of 10m at some point. this is the first thing we need to know