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?

Answer 1

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)

Answer 2

I set the two position equations together and got 0.86 s as my time. Thanks!