Acatapult is designed to launch circus performers from a raised platform. after launch, the height of the performer in feet is given by h (t )equals minus 16 t squared plus 80 t plus 32 where t is seconds after launch. after how many seconds is the performer at ground level? round to the nearest tenth of a second.

x₁  =  5,37 sec

Step-by-step explanation:

The equation of  h(t)  =  -16t²   + 80t   +  32  is in fact one of the equation to describe projectile shot movement  ( in case of shooting is above ground)

y (t)  =  y(o)  + V(o) sin α * t  - g*t²/2    h(t)  =  -16t²   + 80t   +  32

(you can identify by simple  inspection  each term in both equations

To determine after how many seconds is the performer at ground we proceed as follow

h(t)  =  -16t²   + 80t   +  32          at ground level  y  =  h  =  0

Simplifying we get

t²  -  5t   -  2    =  0

A second degree equation, solving for x

x₁,₂  =  [ 5 ± √25 + 8 ] /2            x₁ =  ( 5  + 5,744)/2  and  x₂ =   ( 5  + 5,744)/2

x₂ < 0  we dismiss that root thereis not a negative time

x₁  =  5,372             x₁  =  5,37 sec

inch

step-by-step explanation:

yard and mile are too big.