An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object

vi = 14.610

Explanation:

initial velocity of the first object (vi) = 0 m/s, because it was dropped

distance (y) = -20 m

initial position (y0) = 0 m

acceleration due to gravity (g) = -9.8 m/s^2

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)t^2 + 0 + 0

-20 = -4.9t^2

4.081 = t^2

+√4.081 = t

t = 2.020

time of second object = 2.020 - 1 = 1.020

Now we can plug in the new time to solve for vi of the second object.

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0

-20 = -5.098 + 1.020vi

14.902 = 1.020vi

vi = 14.610 m/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be