# Motion with constant accln help.

1. Apr 6, 2005

### misogynisticfeminist

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 $$s=ut-\frac{1}{2} gt^2$$ substituting, i get,

$$70=84t-1/2 gt^2$$

(are my signs correct)?

$$t= 0.87837 , t= 16.25286236$$

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.

2. Apr 6, 2005

### HallsofIvy

Staff Emeritus
No, you read the problem wrong!

The problem does not ask for the time when the rocket is 70 feet high. It asks for the time interval BETWEEN the two times! Subtract!

3. Apr 6, 2005

### misogynisticfeminist

OH ! i gettit, so i just take away the 2 answers i got. Thanks alot !