Question

You kick a soccer ball with a speed of 18 m/s at an angle of 43°. How long does it take the ball to reach the top of its trajectory?

Answer 1

v=gt
v is vertical velocity, g=9.8 Gravitational acceleration
t=18*sin43°/9.8=1.253s

Answer 2

It helps to realize that the soccer ball has an upward component of velocity (caused by the kick) and a downward component of velocity (caused by the gravity).

The upward component caused by the kick is as follows:\[v _{up} = v _{kick}.\sin \alpha = 18.\sin 43 = 12.3 m/s\]The downward component of velocity is zero to begin with but it grows steadily with time. Gravitational acceleration causes it to increase by 10 m/s with each second passed, so this downward component is 0 m/s at time = 0 sec, then 5 m/s at time = 0.5 sec, then it's 10 m/s at time = 1 sec, and so on. At any given time you can express this component as: \[v _{down} = g.t = (10m/s ^{2}).t\]As you can see from these examples, at first the upward component (which is constant, at 12.3 m/s) is larger than the downward component, so the net vertical motion is upwards. But as the downward velocity increases with time, there comes a moment when the upward velocity and the downward velocity become equal:\[v _{kick}.\sin \alpha = g.t\]At this moment the ball stalls. From this moment on, the downward velocity prevails and the ball goes down.
So the moment when the ball stalls (when it's at the top of its trajectory) occurs at time t:\[t = \frac{ v _{kick}.\sin \alpha }{ g} = \frac{ 12.3 m/s }{10 m/s ^{2} } = 1.23 s\]