Question

The formula
h
=
v
t
−
16
t
2
gives the height
h
, in feet, of an object projected into the air with an initial vertical velocity
v
, in feet per second, after
t
seconds.

If an object is projected upward with an initial velocity of
80
feet per second, at what times will it reach a height of
96
feet above the ground?

It'll be at
96
feet above the ground at
and
seconds

Answer 1

Answer:

t=6

Step-by-step explanation:

We are solving for $t$. To do this, let's fill in the known variables.

$h=vt-16t^2$

$v=80\\h=96$

$96=80t-16t^2$

$16t^2-80t+96=0\\16(t^2-5t+6)=0$

Since the product must equal 0, that means that the factor in the parentheses hasto be equal to 0. Therefore,

$t^2-5t+6=0\\(t+1)(t-6)=0\\\\t+1=0 OR t-6=0\\t=-1 OR t=6$

However, since the time cannot be negative, $t=-1$ cannot be a valid answer.

Therefore, the only valid time is t=6.

I hope this helps.

Answer 2

Let's start with the projectile motion formula which is h = vt - 16t².

In this formula, h is the height of the object,

v is the initial velocity, and t is the time.

We know that the initial velocity is 80 feet per second

so we can plug an 80 in for v in our formula.

We also know that the object reaches a height of 96 feet above the

ground which means we can plug an 96 in for h in our formula.

So we have 96 = 88t - 16t².

To solve this equation, we set it equal to 0 by subtracting

88t and adding 16t² to get 16t² - 88t + 96 = 0².

Now we can divide both sides by 8 to get 2t² - 11t + 12 = 0.

Now we can factor the left side to get (2t - 3)(t - 4) = 0.

So this means that 2t - 3 = 0 or t - 4 = 0.

So t = 3/2 or t = 4.

Now, the reason there are two answer is because the object will be

96 feet off the ground on its way up and on its way down.