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

2.45 seconds

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=(-11\pm√(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.