An object is dropped from a bridge.A second object is thrown downwards 2.00s later.They both reach the water 35.0m below at the same instant.What is the initial speed of the second object?

Question

I will be very grateful if someone could help me out
Heres what I have so far
d=1/2(at^2)
d=35m
35=4.9t^2
t^2=(35/4.9)=7.14
t=2.67 sec
now the second object is thrown 2 sec later.It meets the first object at 2.67-2.00=0.67 sec
The distance travelled by the second object =1/2gt^2=1/2(9.8)(0.67^2)=2.2meters.
the distance travelled by the second object is 35-2.2=32.8 m
so initial velocity=(32.8/0.67)=49m/s

Answer 1

First Stone:
d = 0.5g*t^2 = 35 m.
4.9t^2 = 35
t^2 = 7.14
t = 2.67 s

Second Stone:
t = 2.67-2.0 = 0.67 s. to travel 35 m.
d = Vo*t + 0.5g*t^2
d = Vo*0.67 + 4.9*0.67^2 = 35
Vo*0.67 = 35 - 2.2 = 32.8
Vo = 49 m

Answer 2

Correction: Vo = 49 m/s.