A ball is thrown downward from the top of a 120-foot building with an initial velocity of 14 feet per second. The hoight of the ball h after t

seconds is given by the equation h= - 16 - 14 +120

How long after the ball is thrown will it strike the ground?

Correct question:

A ball is thrown downward from the top of a 120-foot building with an initial velocity of 14 feet per second. The hoight of the ball h after t seconds is given by the equation

h= -16t² -14t +120

How long after the ball is thrown will it strike the ground?

2.34 seconds

Step-by-step explanation:

Given:

Height of building = 120 ft

Initial velocity, u = 14 ft/sec

h = -16t² -14t +120

h(t) = -16t² -14t +120

Let's find t when h(t) = 0

Let's factorize the equation,

-2(8t² + 7t - 60) = 0

We now have a quadartic equation:

8t² + 7 - 60 = 0

This equation cannot be factorized easily, so let's take the quadartic formula:

where a=8, b=7, c=-60

Substituting figures, we have:

Since we can't have a negitive sign, let's take the "+" sign

t = 2.34

The ball will hit the ground at 2.34 seconds

See attached

Step-by-step explanation:

Attached graphs for 1, 3, 4, 8 and 9

Intersections marked on the graph, feel free to chose any additional point