Question 138056
A rocket is launched from atop a 102-foot cliff with an initial velocity of 158
 ft/s. Substitute the values into the vertical motion formula
 h = -16t2 + vt + c. Let h = 0. Use the quadratic formula to find out how long
 after the rocket is launched it will hit the ground. Round to the nearest tenth of a second.
:
Using the values given, the equation will be:
-16t^2 + 158t + 102 = 0
:
the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem a = -16; b=158; c=102
{{{t = (-158 +- sqrt(158^2 - 4 * -16 * 102 ))/(2*-16) }}}
:
{{{t = (-158 +- sqrt(24964 - (-6528 )))/(-32) }}}
:
{{{t = (-158 +- sqrt(24964 + 6528 ))/(-32) }}}
;
{{{t = (-158 +- sqrt(31492))/(-32) }}}
:
You can do the math here, you should get a positive solution of t = 10.483 sec