After grabbing some food from the cafeteria, Ray

gg348892365

gg348892365

Answered question

2022-07-11

After grabbing some food from the cafeteria, Ray trips causing his food to go up in the air before hitting the ground. The parabolic path of his food tray can be modeled by the function ℎ(𝑡) = − 0. 2𝑡 , where h is 2 + 0. 4𝑡 + 1. 6 the height in meters after t seconds. a) What is the maximum height of the tray?

If Ray manages to catch the tray at the same height from which the tray flew out of his hands, how long is it in the air?

Answer & Explanation

Mr Solver

Mr Solver

Skilled2023-06-05Added 147 answers

To find the maximum height of the tray, we need to determine the vertex of the parabolic path. The vertex of a parabola in the form of h(t)=at2+bt+c can be found using the formula t=b2a.
In this case, the equation for the height of the tray is given by h(t)=0.2t2+0.4t+1.6. Comparing this equation with the standard form h(t)=at2+bt+c, we can see that a=0.2, b=0.4, and c=1.6.
Using the formula for the vertex, we have:
t=0.42(0.2)=0.40.4=1
Therefore, the tray reaches its maximum height at t=1 second.
To find the maximum height, we substitute this value of t back into the equation for the height:
h(1)=0.2(1)2+0.4(1)+1.6=0.2+0.4+1.6=1.8
Thus, the maximum height of the tray is 1.8 meters.
If Ray catches the tray at the same height from which it flew out of his hands, it means the height of the tray is zero when it is caught. We can set h(t)=0 and solve for t:
0.2t2+0.4t+1.6=0
This quadratic equation can be solved using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t=b±b24ac2a
Substituting the values a=0.2, b=0.4, and c=1.6:
t=0.4±(0.4)24(0.2)(1.6)2(0.2)
Simplifying:
t=0.4±0.16+1.280.4
t=0.4±1.440.4
t=0.4±1.20.4
Solving for both solutions:
1. When t=0.4+1.20.4=0.80.4=2 seconds.
2. When t=0.41.20.4=1.60.4=4 seconds.
Therefore, the tray is in the air for 2 seconds before Ray catches it.

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