A ball is thrown from a height of

Question

105
feet with an initial downward velocity of
9/fts
. The ball's height
h
(in feet) after
t
seconds is given by the following.
=h−105−9t16t2
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

h(t) = 105 - 9t - 16t^2

When does it hit the ground? That would be when h(t) = 0, right?

So, just solve the quadratic equation.

Answer 2

You mean
g = - 32 ft/s^2
so
h = Hi + Vi t + (1/2)a t^2

h = 105 - 9 t - 16 t^2
at ground:
0 = 105 - 9 t - 16 t^2
or
16 t^2 + 9 t - 105 = 0

t = [ -9 +/- sqrt(81+6720)]/32

t = [-9 +/- 82.5 ]/32
use positive time
2.30 seconds

Answer 3

2.67