- #1
misogynisticfeminist
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This question is the first to pop out in the problems I'm supposed to do and I already have problems with it, so there must be a small little detail which i missed.
A missle is fired vertically upward with an initial velocity of 84m/s from a pt. level with a foot of a tower 70m high. Show that the time from when the missle is first level with the top of the tower until it is again level with the tower a 2nd time to be 15.4 seconds.
What i did was use [tex] s=ut-\frac{1}{2} gt^2[/tex] substituting, i get,
[tex] 70=84t-1/2 gt^2 [/tex]
(are my signs correct)?
this is a quadratic equation and i get answers for time
[tex] t= 0.87837 , t= 16.25286236 [/tex]
the 2nd answer is greater than the given answer of 15.4 seconds. and the first is way too small. I got a feeling i set up the equation wrong. Can anyone help?
thanks.
A missle is fired vertically upward with an initial velocity of 84m/s from a pt. level with a foot of a tower 70m high. Show that the time from when the missle is first level with the top of the tower until it is again level with the tower a 2nd time to be 15.4 seconds.
What i did was use [tex] s=ut-\frac{1}{2} gt^2[/tex] substituting, i get,
[tex] 70=84t-1/2 gt^2 [/tex]
(are my signs correct)?
this is a quadratic equation and i get answers for time
[tex] t= 0.87837 , t= 16.25286236 [/tex]
the 2nd answer is greater than the given answer of 15.4 seconds. and the first is way too small. I got a feeling i set up the equation wrong. Can anyone help?
thanks.