find maximum height for 2nd child, initial speed of 1st child & how long 1st child was in the air?

Question

2 children are playing on 2 trampolines. 1st child can bounce up 1/2 times higher than 2nd child. initial speed up of 2nd child is 5.0 m/s.

Answer 1

Using v0=5, find the height of child 2
using v(t)=5-g*t
at apogee, v(t)=0,
so
t=.5
(g=10)

the height is
y(.5)=5*.5-.5*10*.5^2
1.25 meters

so child 1 rises to 0.625 meters

the time this occurs is
recall
0=v0-g*t at apogee
and
t=v0/g
so
0.625=v0^2/g-.5*v0^2/g
or
v0=sqrt(0.625*2*10)
v0=3.5355 m/s

The time is
0.35355 seconds

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