let h be the height of the building
and hf be the base, or 0
**hf=hi/2 + v'*.55+1/2 g .55^2
but v' is the velocity after it fell the first half
v'=g* timefirsthalf and
hi/2=1/2 g timefirsthalf^2
or timefirsthalf=sqrt (hi/g)
putting this into the equation **,
0=hi/2+.55(g*sqrt(hi/g))+4.86*.55*2
solve for hi. It appears to be a quadratic, I think to simplify it I would substutute x=sqrt(hi) and solve for x with the quadratic equation, then find hi.
A rock dropped from the top of a building takes 0.55 s to fall the last 50% of the distance from the top to the ground.How tall is the building?
4 answers
17
idont understand
Bob Pursley Said:
**hf=hi/2 + v'*.55+1/2 g .55^2
but v' is the velocity after it fell the first half
v'=g* timefirsthalf and
hi/2=1/2 g timefirsthalf^2
or timefirsthalf=sqrt (hi/g)
putting this into the equation **,
0=hi/2+.55(g*sqrt(hi/g))+4.86*.55*2
and set hf to zero but we are adding hi/2 + h(second half) and equation should be:
hi=hi/2+.55(g*sqrt(hi/g))+4.905*.55^2
and when you solve for x you get x = 17.2960032
**hf=hi/2 + v'*.55+1/2 g .55^2
but v' is the velocity after it fell the first half
v'=g* timefirsthalf and
hi/2=1/2 g timefirsthalf^2
or timefirsthalf=sqrt (hi/g)
putting this into the equation **,
0=hi/2+.55(g*sqrt(hi/g))+4.86*.55*2
and set hf to zero but we are adding hi/2 + h(second half) and equation should be:
hi=hi/2+.55(g*sqrt(hi/g))+4.905*.55^2
and when you solve for x you get x = 17.2960032
1 answer