A stone is dropped into a river from a bridge 45.4 m above the water. Another stone is thrown vertically down 1.14 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answer 1

d1 = 0.5g*t^2 = 4.9*1.14^2 = 6.37 m. head start.

h = 0.5g*t^2 = 45.4 m.
4.9t^2 = 45.4
t^2 = 9.27
Tf = 3.04 s. = Fall time of both stones

d2 = d1+ 6.37 m.
Vo*t + 0.5g*t^2 = 0.5g*t^2 + 6.37
Vo*t + 0.5g*t^2 -0.5g*t^2 = 6.37
Vo*t = 6.37
Vo*3.04 = 6.37
Vo = 2.09 m/s.

NOTE: The 2 Free-fall Eqs cancel.

Answer 2

Correction:
d1=0.5g*t^2=4.9*1.14^2=6.37 m Head start.

V = Vo + g*t = 0 + 9.8*1.14=11.17m/s. =
Velocity after falling 6.37 m.

h1 = Vo*t + 0.5g.t^2 =45.4-6.37
11.17t + 4.9t^2 = 39
4.9t^2 + 11.17t - 39 = 0
Tf = 1.9 s. = Fall time of both stones.

h2 = Vo*t + 0.5g*t^2 = 45.4
1.9Vo + 4.9*1.9^2 = 45.4
1.9Vo = 45.4-17.7 = 27.7
Vo = 14.6 m/s. = Initial velocity of 2nd
stone.