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

Answer:

2.11 sec

Step-by-step 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

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

$t=\frac{-4+\sqrt{4^2-4\times 16\times(-80)} }{2\times 16}$

$t=\frac{-4+71.66}{32}$

$t=2.11$