Question

You throw a baseball directly upward at time t=0 at an initial speed of 13.1 m/s. What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take g=9.80 m/s2.
maximum height:

At what times does the ball pass through half the maximum height?

earlier time at half maximum height:

later time at half maximum height:

Answer 1

h = -4.90 t^2 + 13.1 t

t-max is on the axis of symmetry ... t = -b / 2a = -13.1 / (2 * -4.90)
... plug the time into the equation to find the max height

plug half the max height into the equation to find the times (up and down)