A ball is thrown upward from the ground with an initial speed of 26.2 m/s; at the same instant a ball is dropped from rest from a building 14 m high. After how long will the balls be at the same height?

Answer 1

d1 + d2 = 14 m,
26.2 m/s * t + 0.5 * 9.8 m/s^2 *t^2 = 14 m,
26.2t + 4.9t^2 = 14,
4.9t^2 + 26.2t - 14 = 0,
Use quadratic formula to solve for t:

t = (-26.2 +- sqrt(686.44 + 274.4)) /9.8,
t = (-26.2 +- 31) / 9.8
t = 0.49 s or -5.84 s, select + t:

t = 0.49 s.