Aball is thrown vertically in the air with a velocity of 160ft/s. use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 384ft.

2 seconds

Step-by-step explanation:

You are given an equation, a height, and a velocity. Using this you need to solve for t:

We get 2 answers for t. This is because the ball is at 384ft twice - once on its way up, and again on its way down. The ball is at (or above) 384ft for 6 - 4 = 2 seconds.

4 and 6 seconds

Step-by-step explanation:

We have the following information about the problem:

initial velocity:

height:

And the projectile formula is:

substituting known values

To simplify the equation we divide both sides by 16:

Now, we move all terms to the left:

And we have a quadratic equation for the time that can be solved by factoring.

To factor we open two parenthesis and put o each, and we look for two numbers that multiplied result in   and added together result in  . Those numbers are and (because (-4)(-6)=24 and -4+(-6)=-10)

So the factorization is as follows:

and by the zero product property (if two terms when multiplied result in zero, one of them or both are equal to zero):

⇒

⇒

The times that the ball is at a height of 384 ft are 4 and 6 seconds.

.

step-by-step explanation:

.

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic