Question

A ball 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.

Answer 1

Answer:

4 and 6 seconds

Step-by-step explanation:

We have the following information about the problem:

initial velocity: $v_{0}=160ft/s$

height: $h=384ft$

And the projectile formula is:

$h=-16t^2+v_{0}t$

substituting known values

$384=-16ft^2+160t$

To simplify the equation we divide both sides by 16:

$\frac{384}{16} =\frac{-16}{16}t^2+\frac{160}{16}t$

$24=-t^2+10t$

Now, we move all terms to the left:

$t^2-10t+24=0$

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

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

So the factorization is as follows:

$(t-4)(t-6)=0$

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

$(t-4)=0$

⇒ $t=4$

$(t-6)=0$

⇒ $t=6$

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

Answer 2

Answer:

2 seconds

Step-by-step explanation:

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

$384=-16t^2+160t\\16t^2-160t+384=0\\t^2-10t+24=0\\(t-6)(t-4)=0\\t=4, 6$

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.