A projectile was launched from ground level with an initial velocity of the v0 neglecting air resistance is high in feet two seconds after launch is given by S= -16T^2 + v0t find the times that the projectile will reach a high of 128 feet and return to the ground when the zero equals 96 feet per second

what does "the zero equals 96 feet per second" mean?

I think you meant to say v0 = 96 ft/s
s = -16t^2 + 96t
so you need to solve
-16t^2 + 96t = 128
t^2 - 6t + 8 = 0
(t-2)(t-4) = 0
t=2 (on the way up)
t=4 (on the way back down)
and
-16t^2 + 96t = 0
-16t(t-6) = 0
t=0 (at the start)
t=6 (upon impact)

Ah, you're asking about projectile motion! Well, let's dive into it and see if we can solve this conundrum!

We have the equation S = -16T^2 + v0t, where S represents the height of the projectile, T represents time, and v0 represents the initial velocity.

First, let's find the times when the projectile reaches a height of 128 feet. We can set S = 128 and solve for T:

128 = -16T^2 + v0T

Now, if I must be serious for a moment, unfortunately, you haven't provided the value of v0, the initial velocity. So I'm sorry, but I can't solve this equation without that information. But fear not! We'll move on to the second part.

Next, we need to find the times when the projectile returns to the ground at a velocity of 96 feet per second. We can set S = 0 (since the projectile is at ground level) and solve for T:

0 = -16T^2 + 96T

Now, let me tell you a little secret about quadratic equations. Instead of using humor, let's solve this equation using the quadratic formula to find the values of T. The quadratic formula is:

T = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = -16, b = 96, and c = 0. Plugging in these values, we get:

T = (-96 ± √(96^2 - 4(-16)(0))) / (2(-16))

Simplifying further:

T = (-96 ± √(9216)) / -32

T = (-96 ± 96) / -32

Now, here's the funny part! When we simplify, we get:

T1 = (96 - 96) / -32 = 0 / -32 = 0
T2 = (-96 - 96) / -32 = -192 / -32 = 6

So, the projectile will return to the ground at two different times - at t = 0 seconds and t = 6 seconds, assuming our calculations are correct.

Now, remember, without the value for v0, we can't solve for the specific times when the projectile reaches a height of 128 feet. But hopefully, this clears things up a bit, even if it didn't provide the exact answer you were looking for!

To find the times when the projectile reaches a height of 128 feet and returns to the ground, we can set the equation for the height (S) equal to 128 and solve for time (T).

Using the given equation for height: S = -16T^2 + v0T

When the projectile reaches a height of 128 feet:
128 = -16T^2 + v0T

Now, let's substitute v0 = 96 (given) into the equation:
128 = -16T^2 + 96T

To solve this quadratic equation, let's rearrange it into standard form:
-16T^2 + 96T - 128 = 0

Now we can either factor, complete the square, or use the quadratic formula to solve for T. Let's use the quadratic formula:

T = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -16, b = 96, and c = -128. Substituting these values into the quadratic formula:

T = (-96 ± √(96^2 - 4*(-16)*(-128))) / (2*(-16))

T = (-96 ± √(9216 - 8192)) / (-32)

T = (-96 ± √1024) / (-32)

T = (-96 ± 32) / (-32)

Now, we can solve for T:

T1 = (-96 + 32) / (-32) = -64 / (-32) = 2 seconds
T2 = (-96 - 32) / (-32) = -128 / (-32) = 4 seconds

Therefore, the projectile will reach a height of 128 feet and return to the ground at 2 seconds and 4 seconds after launch.

To find the times at which the projectile reaches a height of 128 feet and returns to the ground, we can use the equation for height in terms of time: S = -16t^2 + v0t.

1. Reaching a height of 128 feet:
We want to find the time(s) at which the height (S) equals 128 feet. To do this, substitute S = 128 into the equation:
128 = -16t^2 + v0t.

We need the initial velocity (v0) to continue. In this case, you mentioned that v0 equals 96 feet per second, so substitute v0 = 96 into the equation to get:
128 = -16t^2 + 96t.

Now you have a quadratic equation. Rearrange it to standard form:
-16t^2 + 96t - 128 = 0.

To solve this quadratic equation for t, you can factor it, complete the square, or use the quadratic formula. Whichever method you choose, you'll find two values for t that correspond to the projectile reaching a height of 128 feet.

2. Returning to the ground:
To determine when the projectile returns to the ground, we need to find the time(s) when the height (S) equals zero. Substitute S = 0 into the equation:
0 = -16t^2 + 96t.

Again, we have a quadratic equation. Rearrange it to standard form:
-16t^2 + 96t = 0.

Factor out common terms to simplify:
-16t(t - 6) = 0.

Set each factor equal to zero to find t:
-16t = 0 (giving t = 0),
or (t - 6) = 0 (giving t = 6).

So, the projectile returns to the ground at t = 0 seconds and t = 6 seconds.

By solving the first equation, you will find the times at which the projectile reaches a height of 128 feet, and by solving the second equation, you will find the times at which the projectile returns to the ground.