A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.87 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

S=ut-(1/2)gt²
S=41.7 m, u=0, g=9.8 m/s²
Solve for t.
Now
S=v0(t-1.87) - (1/2)g(t-1.87)²
Solve for v0.