Question

A ball is thrown from a height of 80 feet with an initial downward velocity of 4/fts. The ball's height h (in feet) after t seconds is given by the following.

h=80-4t-16t^2
How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth.

Answer 1

Alright, so we have h=80-4t-16t^2. We have to find when h=0. We can use the quadratic formula

(use x as t)

x=(4+-sqrt(16-(-64*80))/32=(4+-sqrt(5136))/(-32) , and since 4+sqrt(5136)/(-32) is clearly negative that doesn't work, so we have (4-sqrt(5136))/(-32), which is approximately 2.11

Answer 2

2.11 sec

Explanation:

Given that the ball is thrown from a height of 80 feet

it follows the equation

`h = 80-4t-16t^2`

when the ball hits the ground the height if the ball is equal to 0

`16t^2+4t-80=0`

`t=(-4+√(4^2-4* 16*(-80)) )/(2* 16)`

`t=(-4+71.66)/(32)`

`t=2.11`