An object is released into the air at an initial height of 6 feet and an initial velocity of 30 feet per second. The object is caught at a height of 7 feet. Use the vertical motion model, h = -16t^2+vt+s, where h in the height, t is the time in motion, s is the initial height, and v is the initial velocity, to find how long the object is in motion.

t=1.8125

t=-7/2 1.84 per sec

To find the time the object is in motion, we need to solve the equation for t in the vertical motion model.

The vertical motion model is given by:
h = -16t^2 + vt + s

Substituting the given values, we have:
h = -16t^2 + 30t + 6
Since we know that the object is caught at a height of 7 feet, we can substitute h with 7:
7 = -16t^2 + 30t + 6

Rearranging the equation, we have:
-16t^2 + 30t - 1 = 0

To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = -16, b = 30, and c = -1.
Plugging in these values into the quadratic formula, we get:
t = (-30 ± √(30^2 - 4(-16)(-1))) / (2(-16))

Simplifying further:
t = (-30 ± √(900 - 64)) / (-32)
t = (-30 ± √(836)) / (-32)

Taking the square root of 836:
t = (-30 ± √(4 * 209)) / (-32)
t = (-30 ± √(4) * √(209)) / (-32)
t = (-30 ± 2√(209)) / (-32)

Now, we have two possibilities:
1. Taking the positive square root:
t = (-30 + 2√(209)) / (-32)

2. Taking the negative square root:
t = (-30 - 2√(209)) / (-32)

Therefore, the object is in motion for either of the two calculated values of t.

To find how long the object is in motion, we need to solve the given vertical motion model equation for time (t).

The vertical motion model equation is:
h = -16t^2 + vt + s

Given:
s = 6 feet (initial height)
v = 30 feet per second (initial velocity)
h = 7 feet (height where the object is caught)

Substituting the given values into the equation, we get:
7 = -16t^2 + 30t + 6

Now, we have a quadratic equation. We can rearrange it to standard form:
-16t^2 + 30t + 6 - 7 = 0
-16t^2 + 30t - 1 = 0

To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, our quadratic equation is in the form of at^2 + bt + c = 0, where:
a = -16
b = 30
c = -1

Applying the values to the quadratic formula, we get:
t = (-30 ± sqrt(30^2 - 4(-16)(-1))) / 2(-16)
t = (-30 ± sqrt(900 - 64)) / -32
t = (-30 ± sqrt(836)) / -32
t = (-30 ± 28.913) / -32

Now, we'll calculate both possibilities:
t1 = (-30 + 28.913) / -32
t1 = -1.087 / -32
t1 = 0.034 seconds (approximately)

t2 = (-30 - 28.913) / -32
t2 = -58.913 / -32
t2 = 1.843 seconds (approximately)

So, there are two possible solutions for time (t): t ≈ 0.034 seconds and t ≈ 1.843 seconds.

Therefore, the object is in motion for approximately 0.034 seconds and 1.843 seconds.