Algebra equation

Question

H(t)=-16t²+224t+28.25

 

distance is d(t)=31.7t

The height (h) of the football in feet (t) seconds after it is thrown is modeled by h(t)=-16t^2+224t+28.25 .The horizontal distance (d) in feet between the quarterback and the football is modeled by d(t)=31.7t. To the nearest foot How far does the football land from the quarterback?

Answer 1

The first thing you have to figure out is how long the ball is in the air. We know it'll be in the air until the height of the ball is 0.

 

h(t) = 0 = -16t^2 + 224t + 28.25

 

t = (-b + sqrt(b^2-4*a*c))/2a

 

a = -16

b = 224

c = 28.25

 

t = (-224 + sqrt(224^2-4*(-16)*28.25))/(2*(-16))

 

= (-224 + sqrt(50176+1808))/(-32)

 

= (-224 + 228)/(-32)

 

t = 4/(-32), -452/-32

 

t = -1/8, 113/8

 

Since time can't be negative, we disregard the first answer.

 

t = 113/8, which is about 14 seconds of airtime. It'll be traveling according to our distance equation throughout the duration of its airtime.

 

D = 31.7*t

 

D = 31.7*113/8 = 448 feet, just short of 1 and a half football fields.

 

That is a crazy strong quarterback.