Question

Consider the quadratic model h(t)=-16t^2+40t+50 for the height (in feet), h, of an object t seconds after the object has been projected straight up into the air. Find the maximum height attained by the object. How much time does it take to fall back to the ground? Assume that it takes the same time for going up and coming down.

Answer 1

To find the maximum height reached by the object, calculate the vertex of the quadratic equation, resulting in a height of 75 feet at t=1.25 seconds. To find the time to fall back to ground, set h(t)=0 and solve for t, yielding 2.5 seconds in total if air resistance is ignored.

Step-by-step explanation:

The given quadratic model h(t)=-16t^2+40t+50 represents the height in feet of an object t seconds after being projected straight up into the air. To find the maximum height attained by the object, we need to find the vertex of the parabola because this is where the maximum height will be. The vertex formula for a parabola in the form of y=ax^2+bx+c is h=-b/(2a). Using this formula, we can find the time it takes to reach the maximum height (tmax):

tmax = -40 / (2 * -16) = 1.25 seconds

Plugging tmax back into the quadratic equation:

h(1.25) = -16(1.25)^2 + 40(1.25) + 50 = 75 feet.

So the maximum height reached by the object is 75 feet. To find the time it takes to fall back to the ground, we set h(t) equal to zero and solve for t:

0 = -16t^2 + 40t + 50

Using the quadratic formula, we find two possible times (t1 and t2), which represent the upward and downward part of the journey. Since the trajectory is symmetrical, the time it takes to go up is equal to the time it goes down. Therefore, the total time will be twice the time to reach the maximum height (1.25 s), unless there is a significant amount of air resistance.

In the absence of air resistance, the object will take 2.5 seconds to fall back to the ground.

Answer 2

The results to the options provided, it seems closest to option D: maximum height 75 feet and time to reach the ground 2.5 seconds.

a. Maximum Height:

We want to find the maximum height attained by the object, which occurs at the vertex of the quadratic function
`\(h(t) = -16t^2 + 40t + 50\).`

The formula for the time at the vertex is
`\(t = -(b)/(2a)\):`

`\[t = -(40)/(2(-16)) = (40)/(32) = 1.25 \text{ seconds}\]`

Now, let's find the height at t = 1.25 seconds:

`\[h(1.25) = -16(1.25)^2 + 40(1.25) + 50\]`

`\[h(1.25) = -16(1.5625) + 50 + 50\]`

`\[h(1.25) = -25 + 100\]`

`\[h(1.25) = 75 \text{ feet}\]`

So, the maximum height attained by the object is indeed 75 feet.

b. Time to Reach Ground:

We'll find the time it takes to reach the ground by setting the height function h(t) equal to zero:

`\[0 = -16t^2 + 40t + 50\]`

Let's solve this quadratic equation to find the time it takes to reach the ground.

Using the quadratic formula
`\(t = (-b \pm √(b^2 - 4ac))/(2a)\):`

`\(a = -16\),\(b = 40\),\(c = 50\)`

`\[t = (-40 \pm √(40^2 - 4(-16)(50)))/(2(-16))\]`

`\[t = (-40 \pm √(1600 + 3200))/(-32)\]`

`\[t = (-40 \pm √(4800))/(-32)\]`

`\[t = (-40 \pm 69.28)/(-32)\]`

The solutions will be
`\(t = (-40 + 69.28)/(-32)\) and \(t = (-40 - 69.28)/(-32)\).`

Calculating these gives:

`\[t = (29.28)/(-32) \approx -0.915 \text{ seconds}\]`

`\[t = (-109.28)/(-32) \approx 3.415 \text{ seconds}\]`

These values don't align with the assumption that the time to reach the ground is equal for the ascent and descent. It's likely that the negative root is an extraneous solution.

Therefore, the time it takes to fall back to the ground is approximately 3.415 seconds.

Complete question: