Asked by Joey
Two balls are dropped, one from height H and the other from a height of 4H. How far apart should they be dropped (time wise) to hit the ground at the same time.
Answers
Answered by
bobpursley
h=1/2 g t^2 is the basic formula.
H=1/2 g (t-to)^2 where to is the time the lower one is dropped, after the higher.
4H=1/2 g t^2
So, both balls hit at time t.
t=sqrt 8H/g
H=1/2 g (sqrt(8H/g)-to)^2
2H=g(8H/g-2to*sqrt(8H/g)+to^2)
so you have a quadratic...
to^2-2to* sqrt(8H/g)-2H/g
a=1 b=-2sqrt(8H/g) c= -2H/g
solve by the quadratic equation.
H=1/2 g (t-to)^2 where to is the time the lower one is dropped, after the higher.
4H=1/2 g t^2
So, both balls hit at time t.
t=sqrt 8H/g
H=1/2 g (sqrt(8H/g)-to)^2
2H=g(8H/g-2to*sqrt(8H/g)+to^2)
so you have a quadratic...
to^2-2to* sqrt(8H/g)-2H/g
a=1 b=-2sqrt(8H/g) c= -2H/g
solve by the quadratic equation.
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