Question

A ball is thrown from a height of
123
feet with an initial downward velocity of
11/fts
. The ball's height
h
(in feet) after
t
seconds is given by the following.
=h−123−11t16t2
How long after the ball is thrown does it hit the ground?

Answer 1

17+x9x13 =428is 5-8h34z

Answer 2

Answer:

2.45 seconds

Step-by-step explanation:

A ball is thrown from a height of  123 ft

Thus, $h_0=123\ ft$

Initial velocity, $u=-11\ ft/s$

Acceleration due to gravity downwards, $g=-32\ ft/s^2$

The ball's height  h  (in feet) after  t  seconds is given by,

$h=-11t-16t^2+123$

When ball hit the ground, h=0

$0=-11t-16t^2+123$

$16t^2+11t-123=0$

Using quadratic formula,

$t=\dfrac{-11\pm\sqrt{11^2-4(16)(-123)}}{2(16)}$

$t=-3.14,2.45$

Time can't be negative.

So, t=2.45 seconds

Hence, After 2.45 seconds ball hit the ground.