Answer:
[tex]h=9.30m[/tex]
Explanation:
We have an uniformly accelerated motion, due to the gravitational acceleration. So, we use the kinematic equations, since the ball is throw directly upward, g is negative:
[tex]h=v_0t-\frac{gt^2}{2}[/tex]
First, we need to calculate the time taken by the ball to reach the maximum height, in this point its final speed is zero:
[tex]v_f=v_0-gt\\\\\frac{0-v_0}{-g}=t\\t=\frac{v_0}{g}\\t=\frac{13.5\frac{m}{s}}{9.8\frac{m}{s^2}}\\t=1.38s[/tex]
Now, we can calculate h:
[tex]h=v_0t-\frac{gt^2}{2}\\h=13.5\frac{m}{s}(1.38s)-\frac{9.8\frac{m}{s^2}(1.38s)^2}{2}\\h=9.30m[/tex]
