Question

A football is thrown with an initial upward velocity of 25 feet per second from a height of 5 feet above the ground. The equation h = −16t^2 +25t + 5 models the height in feet t seconds after it is thrown. After the ball passes its maximum height, it comes down and hits the ground. About how long after it was thrown does it hit the ground?

Answer 1

A = -16
B = 25
C = 5

Plug into the quadratic equation and you get: -0.18 and 1.75. Since we cannot have a negative time our answer is 1.75 seconds.

Answer 2

Answer:

1.74 second long after it was thrown does it hit the ground.

Step-by-step explanation:

Given : A football is thrown with an initial upward velocity of 25 feet per second from a height of 5 feet above the ground. The equation $h =-16t^2+25t+5$ models the height in feet t seconds after it is thrown.

To find : How long after it was thrown does it hit the ground?

Solution :

The equation model is $h(t)=-16t^2+25t+5$

After the ball passes its maximum height, it comes down and hits the ground.

i.e. h=0

So, $-16t^2+25t+5=0$

Solve by quadratic formula of equation $ax^2+bx+c=0$ is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=25 and c=5

$t=\frac{-25 \pm \sqrt{(25)^2-4(-16)(5)}}{2(-16)}$

$t=\frac{-25 \pm \sqrt{625+320}}{-32}$

$t=\frac{-25 \pm \sqrt{945}}{-32}$

$t=\frac{-25+\sqrt{945}}{-32},\frac{-25-\sqrt{945}}{-32}$

$t=-0.179,1.741$

We reject t=-0.179.

Therefore, 1.74 second long after it was thrown does it hit the ground.