Posted by <3 on Saturday, September 14, 2013 at 2:09am.
A ball is thrown upward from the ground with an initial speed of 23.2 m/s; at the same instant, another ball is dropped from a building 20 m high. After how long will the balls be at the same height?

Physics  Graham, Saturday, September 14, 2013 at 4:49am
Given:
Let the first ball have height x, velocity u.
x(0) = 0[m]
u(0) = 23.2[m/s]
x(t) = x(0) + u(0) t + (g/2) t^2
.: x(t) = 23.2 t  4.9 t^2
Let the second have height y, velocity v.
y(0) = 20[m]
v(0) = 0[m/s]
y(t) = y(0) + v(0) t + (g/2) t^2
.: y(t) = 20  4.9 t^2
Where: g = 9.8[m/s^2]
Find: t such that x(t)=y(t)

Physics  <3, Sunday, September 15, 2013 at 10:50pm
I set the two position equations together and got 0.86 s as my time. Thanks!
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