A stone is dropped into a river from a bridge 43.3 m above the water. Another stone is thrown vertically down 1.06 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

h=gt²/2 =>
t= sqrt(2h/g) = sqrt(2•43.3/9.8) =2.97 s.
t1=t-1.06 = 2.97 – 1.06 = 1.91 s.
h= v₀•t1 +g•t1²/2
Solve for v₀.