Question

A ball is thrown at an angle of 30° off the horizontal, with an initialvelocity of 28 m/s. What is the maximum height the ball will reach?

Answer 1

$sen(30)\ast28=\frac{14m}{s}$$\begin{gathered} v^2=(14m/s)^2-2ad \\ \frac{14^2}{2\ast9.81}=d=9.99m \end{gathered}$