Question 30981
<pre><b>Next time please provide a complete information: Your formula seems to be missing here:
The formula is:
*[tex h(t)=\frac{-1at^2}{2}+vt+s] where a is the acceleration due to gravity,v is the upward speed of the object upon release, and s is the starting height of the object and t is the seconds after being relaeased.
at the surface of the Earth a is *[tex 32ft/s^2]
Equation:
{{{h(t)=-((32)t^2)/2+30t+3}}}
You have to solve for t:
Multiply the whole equation by -2 to remove the fraction and the negative; and simplify:
{{{h(t)=32t^2-60t-6}}}
Use the quadratic formula:{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In which a=32, b=-60, c=-6
{{{x=(-(-60)+-sqrt((-60)^2-4*32*-6))/(2*32)}}}
{{{x=(60+-sqrt(4368))/64}}}
Remove the negative:
{{{x=(60+66)/64}}}
Simplfy:
x=1.97
Letting x=t
t=1.97
<font color = blue>Hence, it tool the rocket 1.97 seconds to come down.</font>
Paul.