Question

A catapult 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.

Answer 1

x₁ = 5,37 sec

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