two balls are juggled so that each ball is in the air for 1.800s, and the second ball is thrown up at the instant the first starts down.
Find the height above the throwing hand where the two balls pass eachother, assuming gEarth=9.800m/s^2
I started with d=vt
but we don't have the velocity of the ball and how do we find the velocity if we don't have the distance?
time for the second starts at (t-.9)
at some points, they have the same height.
H1=vi*t-4.9t^2
H2=vi(t-.9)-4.9(t-.9)^2
set them equal, solve for time t when they are at the same height.
Then, put that t into either equation, and solve for height h.