A ball is thrown vertically upward with a velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. At what position above the ground will they meet?
math - Steve, Monday, July 30, 2012 at 9:24am
you want to find where
18t - 4.9t^2 = 13.5(t-2) - 4.9(t-2)^2
t = 3.086 seconds
height: 8.883 m