Question

You throw a baseball directly upward at time f = 0 at an initial speed of 12.5 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9 80 m/s^2.

Answer 1

at max ht, velocity = 0,

we know the equation as
v² = u² + 2as
0 = 12.5² - 2*9.8s
s = 12.5² /19.6 = 7.972 m

half the max height = 3.986 m
using the equation
s = ut + 0.5at², we get the quadratic
3.986 = 12.5t - 4.9t² , which, on solving, we finally get
t = 0.373592 s or t = 2.17743 s ................................................ans