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.

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step-by-step explanation: