Respuesta :
Answer:
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
