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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