While standing on a bridge 18.0 m above the ground, you drop a stone from rest.

When the stone has fallen 3.10 m, you throw a second stone straight down. What minimal initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

V=____________ m/s

Sorry about my mistake!

Displacement = vi * t + ½ * a * t^2

1st stone
vi = 0
a = -9.8 m/s^2
d = -3.1

-3.1 = ½ * -9.8 * t^2
t^2 = -6.2 ÷ -9.8
t = (-6.2 ÷ -9.8)^0.5 = time to fall 3.1 m

AND

-18 = ½ * -9.8 * t^2
t^2 = -36 ÷ -9.8
t = (-36 ÷ -9.8)^0.5 = time to fall 18 m



The time for the 2nd stone to reach the ground is the difference of the 2 times above, which = (-36 ÷ -9.8)^0.5 – (-6.2 ÷ -9.8)^0.5 = 1.9167 – 0.7954 = 1.1212 s

The 2nd stone must fall 18 m in the time above
-18 = vi * 1.1212 + ½ * -9.8 * 1.1212^2

-18 = vi * 1.1212 + -6.1597
vi = -11.8403 ÷ 1.1212 = -10.56 m/s


Check
-18 = -10.56* 1.1212 + ½ * -9.8 * 1.1212^2

-18 = -11.8399 + -6.1597 = -18

OK

First stone has a headstart given by:
3.1=0.5*9.81*t^2
t=0.795 s
It will hit the ground after time:
18=0.5*9.81*t^2
t=1.916 s
So second stone must hit the ground in 1.916 - 0.795 s
=1.121 s
s=ut+0.5*a*t^2
-18=-u*1.121-0.5*9.81*1.121^2
-18=-u*1.121 - 6.164
u=10.56 m/s