A potato is launched vertically upward with an initial velocity of 34ft/s from a potato gun at the top of a building that is 15 feet tall. The height of the potato in feet after t seconds is given by s(t)=−16t2+34t+15. Determine how long the potato is in the air. Enter an exact answer.

It will be in the air as long as s(t) > 0

So let's see when it hits the ground
-16t^2 + 34t + 15 = 0
16t^2 - 34t - 15 = 0
t = (34 ± √2116)/32
= (34 ± 46)/32 = 2.5 or a negative which must be rejected

so the potato is in the air for 2.5 seconds

Well, let's calculate how long the potato is in the air by finding the time when it reaches the ground. The time it takes for the potato to hit the ground is when the height, s(t), is equal to zero.

So, we set the equation -16t^2 + 34t + 15 equal to zero:

-16t^2 + 34t + 15 = 0

Now, I could solve this using the quadratic formula, but solving it feels a bit too "mashed" for me. Instead, I'll "root" for factoring!

Factoring -16t^2 + 34t + 15 = 0, we have:

(-4t - 3)(4t - 5) = 0

Setting each factor equal to zero:

-4t - 3 = 0 or 4t - 5 = 0

Solving the first equation: -4t = 3 --> t = -3/4

Solving the second equation: 4t = 5 --> t = 5/4

Now, since time cannot be negative, we ignore the negative solution and only consider the positive solution. Therefore, the potato is in the air for t = 5/4 seconds.

So, the potato is in the air for 5/4 seconds, or in spaghetti terms, about "a-mac-ing" 1.25 seconds.

To determine how long the potato is in the air, we need to find the time when the height of the potato is zero, as that indicates that the potato has reached the ground.

The height of the potato at any time t is given by the equation s(t) = -16t^2 + 34t + 15.

To find when the height is zero, we set s(t) to zero and solve for t:

0 = -16t^2 + 34t + 15

Now, let's solve this quadratic equation.

We can start by factoring out -1 from the coefficient of the squared term to simplify the equation:

0 = -1(16t^2 - 34t - 15)

Next, we can factor the quadratic expression inside the parentheses:

0 = -1(2t - 3)(8t + 5)

Setting each factor equal to zero, we have:

2t - 3 = 0 or 8t + 5 = 0

Solving for t in each equation:

For 2t - 3 = 0:
2t = 3
t = 3/2

For 8t + 5 = 0:
8t = -5
t = -5/8

Since time cannot be negative in this context, we discard the solution t = -5/8.

Therefore, the potato is in the air for t = 3/2 seconds.

To determine how long the potato is in the air, we need to find the time when the height of the potato is equal to zero. This will be the time when the potato reaches the ground.

In this case, the height function is given as s(t) = -16t^2 + 34t + 15.

Setting s(t) = 0, we have:

-16t^2 + 34t + 15 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, factoring may not be straightforward, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values from the given equation:

a = -16, b = 34, c = 15

t = (-34 ± √(34^2 - 4(-16)(15))) / (2(-16))

Simplifying further:

t = (-34 ± √(1156 + 960))/(-32)

t = (-34 ± √(2116))/(-32)

t = (-34 ± 46)/(-32)

Now we have two possible values for 't':

t1 = (-34 + 46)/(-32) = 12/(-32) = -3/8

t2 = (-34 - 46)/(-32) = -80/(-32) = 5/2

Since time cannot be negative, we discard the negative value of t.

Therefore, the potato is in the air for 5/2 or 2.5 seconds.