Question

If a football player passes a football from 4 feet off the ground with an initial velocity of 36 ft./s, How long will it take the football to hit the ground? Use the equation H equals -16 T to the 2nd+6t +4

Answer 1

It will take 0.72s for the football to hits the ground.

Explanation:

We have that the equaation for the height of the football is

`H(t) = -16t^(2) + 6t + 4`

The football will hit the ground when
`H(t) = 0`.

`H(t) = -16t^(2) + 6t + 4`

`-16t^(2) + 6t + 4 = 0`

Multiplying by -1

`16t^(2) - 6t - 4 = 0`

To solve this equation, we need the bhaskara formula:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = (x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4a`

In this problem, we have that:

`16t^(2) - 6t - 4 = 0`

So

`a = 16, b = -6, c = -4`

`\bigtriangleup = (-6)^(2) - 4*16*(-4) = 292`

`t_(1) = (-(-6) + √(292))/(2*16) = 0.72`

`t_(2) = (-(-6) - √(292))/(2*16) = -0.35`

It cannot take negative seconds for the ball to hit the ground.

So it will take 0.72s for the football to hits the ground.