free fall physics

Question

A baseball pitcher makes a big mistake and throws the ball straight up. If it reaches a maximum height of 15m find (A) the total time the ball spent in the air (B) the ball's average velocity on the way up to the top (C) the ball's velocities when its displacement was equal to 8.0 m

Answer 1

h(t) = -4.9t^2 + vo(t) where vo is the initial velocity, t= time

h'(t) =-9.8t + vo = 0 at maximum height

vo = 9.8t

15 = -4.9t^2 +vo(t) = -4.9t^2 + 9.8t^2 = 4.9t^2

t^2 = 15/4.9 = 3.06

t= sq3.06 = 1.75 seconds at maximum height

vo =9.8(1.75) = 17.15 meters per sec

double t =1.75 to get total time in the air

t= 1.75(2) = 3.5 seconds

when height = 8

8 = -4.9t^2 + 17.15t

-4.9t^2 + 17.15t -8 = 0

t = 17.15/4.9 + (1/4.9)sqr(17.15^2 - 4(4.9)(8))

3.5 + or - 2.4 = 1.1 or 5.9

t = 1.1 or 5.9 seconds when height = 8

3.5-1.1 = 2.4

5.9-3.5 = 2.4

the maximum height is equidistant in time from the two times the height = 8