A ball falls from the top of a high cliff. A second ball is thrown downward from the same height 4 secs later with an initial speed of 40m/s. If both balls hits the ground simultaneously. How high is the cliff?

Answer 1

0 = Hi - Vi t - 4.9 t^2

0 = Hi - 0 (t+4) - 4.9 (t+4)^2
0 = Hi - 40 t - 4.9 t^2
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Hi = 4.9 t^2 + 40 t
Hi = 4.9(t^2+8t+16)
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4.9 t^2+40 t = 4.9 t^2+39.2t +78.4
.8t = 78.4
t = 98

Hi = 40t+4.9t^2 =40*98+4.9(98^2)
=50,980 meters (30 miles? check arithmetic :)