Question

In American football, the ball is punted by dropping and kicking it before it hits the ground. The height h(t) of the football above the ground in meters t seconds after being punted is affected by gravity and by the punter's kick, and can be represented as the difference of two functions: a(t)=1.4t2, which specifies the effect of gravity on the height of the ball, and v(t)=12.1t+2.5, which specifies the effect of the punt on the ball. With these two functions, h(t)=v(t)-a(t), how high above the ground was the ball when it was punted?

Answer 1

Answer:

h = 8.845 m

Step-by-step explanation:

The height h(t) of the football above the ground in meters t seconds after being punted is affected by gravity and by the punters kick, and can be represented as the difference of two functions:

$a(t)=1.4t^2\\\\v(t)=12.1t+2.5$

We need to find h(t) such that, h(t)=v(t)-a(t)

So,

$h(t)=12.1t+2.5-1.4t^2\\\\h(t)=-1.4t^2+12.1t+2.5$

It is a quadratic equation. When we solve it we get :

$h(t)=8.845\ m, -0.2\ m$

Neglecting negative value,

h(t) = 8.845 m

So, the ball was at a height of 8.845 m when it was punted.