Explanation:
Assuming that there is no air resistance, the given time means that the ball reached its maximum height 3 seconds after throwing it upwards so we can solve its
maximum height by solving for [tex]v_{0y}[/tex] first:
[tex]v_y = v_{0y} - gt \Rightarrow v_{0y} = gt[/tex]
[tex]v_{0y} = (9.8\:\text{m/s}^2)(3\:\text{s}) = 29.4\:\text{m/s}[/tex]
Now that we know [tex]v_{0y},[/tex] we can now solve for [tex]y_{max}:[/tex]
[tex]y_{max} = v_{0y}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:\:\:=(29.4\:\text{m/s})(3\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(3\:\text{s})^2[/tex]
[tex]\:\:\:\:\:\:=44.1\:\text{m}[/tex]
