# The height of a swing above ground can be modelled by equation : h=1.5+sin 3t when is the swing first 1m above the ground(to the nearest one tenth of a second)?

### 3 Answers | Add Yours

The height can be modelled by the following equation.

h=1.5+sin 3t

When the swing is 1m above ground h = 1m

`1 = 1.5+sin3t`

`-0.5 = sin3t`

`-sin(pi/6) = sin3t`

We know that

`sin(pi+theta) = -sintheta `

Let us say;

`sin(pi+pi/6) = sin3t`

`3t = (7pi/6)`

`t = (7pi)/18`

`t = 1^0 13^0 18''`

*Answer to the nearest tenth second;*

`t = 1^0 13^0 20''`

*Note*

*The calculations are considered as motion starting by increment angle.*

Note that the funcytion has the same period T of function y= sin 3t

Indeed: if `3/2+sin3t_1=3/2+sin3t_2 rArr sin 3t_1= sin3t_2`

that is: `t_2= pi-t_1` I one round of circle.

Then :

`t_1=1.221730476396030703846583537942`

`t_2=1.9198621771937625346160598453375`

To know when the swiing is at 1 mt from the ground you have to solve eaquation :

`3/2+sin3t=1`

That means: `sin 3t= 1-3/2=-1/2`

This causes: `3t= 11/6 pi+2kpi` `t=11/18 pi +2/3 kpi`

and: `3t=7/6 pi+2kpi` `t=7/18 pi +2/3 kpi`

So solution are `t_1=1/18 pi (11+12k)` `t_2=1/18 pi (7+12k)`

Let you see, red line is function y= 3/2+ sin3t, while blue line is function y=1

The point you are searching for are the point wich red line intercept blue one.

TO BE CONTINUE.....