Question

A ball is thrown at an angle of 50 degrees above the horizontal. After 5 s, the ball is at a distance 38 m along the horizontal from the point where it was released. How high is the ball above the point where it was released during this time?

Answer 1

As we established previously, the initial velocity is 11.8 m/s. So, the height is

$$H=v\sin\theta ·t-gt^2/2=-77\text{ m}$$

(below the point where it started).