Question 1155173
 A rocket is launched from atop a 73-foot cliff with an initial velocity of 142 ft/s.
 The height of the rocket above the ground at time t is given by h=-16t^2+142t+73.
:
A) When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.(Solve Algebraically, Hint: Use the Quadratic Formula to solve and the round the square root answer one decimal place)
h = 0 when it hits the ground
-16t^2 + 142t + 73 = 0
The quadratic formula: {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=-16, b=142, c=73
{{{t = (-142 +- sqrt(142^2-4*-16*73 ))/(2*-16) }}}
:
{{{t = (-142 +- sqrt(20164+4672 ))/(-32) }}}
:
{{{t = (-142 - sqrt(24836))/-32 )) }}}, the positive solution
do the math
t = 9.3623 ~ 9.4 seconds
:
B) When will the rocket reach 199ft from the ground? (Solve Algebraically)
-16t^2 + 142t + 73 = 199
-16t^2 + 142t + 73 - 199 = 0
-16t^2 + 142t - 126 = 0
a=-16; b=-142, c=-126
{{{t = (-142 +- sqrt(142^2-4*-16*-126 ))/(2*-16) }}}
:
{{{t = (-142 +- sqrt(20164-8064 ))/(-32) }}}
:
{{{t = (-142 - sqrt(12100))/(-32) }}}
do the math
t = 7.875 ~ 7.9 seconds reaches 199 ft the 2nd time
and
{{{t = (-142 + sqrt(12100))/(-32) }}}
t = 1 second reaches 199 ft the first time