Mathematics, 16.11.2020 16:40 johnkings140

The height h, in feet, of objects such as thrown balls can be modeled as a function of time t, in seconds, as h(t)=−16t2+v0t+h0, where the initial upward velocity is v0, in feet per second, and the initial height is h0, in feet. Marla and Kerry are practicing lacrosse tosses. Marla throws a ball from a height of 6 feet with an initial vertical velocity of 40 feet per second. At the same instant, Kerry throws a ball from a height of 5 feet with an initial vertical velocity of 50 feet per second. Which ball is in the air longer and by how many seconds more?

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Answer from: steph76812

Kerry's ball stays longer in the air than Marias ball by an extra 0.582 seconds

Step-by-step explanation:

The equation that models the height, h, in feet of object thrown modeled as a time function is given as follows;

h(t) = -16t² + v₀·t + h₀

Where;

v₀ = The velocity upwards (vertical velocity)

h₀ = The initial height of the object in feet

The given parameters are;

The initial height from which Maria throws a ball = 6 feet

The initial vertical velocity with which Maria throws the ball = 40 feet per second

The initial height from which Kerry throws a ball = 5 feet

The initial vertical velocity with which Kerry throws the ball = 50 feet per second

For Maria, from the equation for the height of the object, we have;

h(t) = -16t² + v₀·t + h₀

Where for Maria, we have;

h₀ = 6 feet

v₀ = 40 ft/s

Substituting gives;

h(t) = -16 × t² + 40 × t + 6 = -16·t² + 40·t + 6

h(t) = -16·t² + 40·t + 6

The time it takes for the ball to go up and come back down again is given by the value of t, when h = 0 (ground level) as follows;

0 = -16·t² + 40·t + 6

16·t² - 40·t - 6 = 0

From the quadratic formula, we have

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

$t = \dfrac{40\pm \sqrt{(-40)^{2}-4\times 16 \times (-6)}}{2\times 16} = \dfrac{40 \pm 8 \times \sqrt{31} }{32} = \dfrac{5}{4} \pm \dfrac{\sqrt{31} }{4}$

t ≈ 2.64 s or -0.14 s

Therefore, the time the ball spends in the air ≈ 2.64 s

For Kerry, we have;

h₀ = 5 feet

v₀ = 50 ft/s

Substituting gives;

h(t) = -16 × t² + 50 × t + 5 = -16·t² + 50·t + 5

h(t) = -16·t² + 50·t + 5

Which gives;

16·t² - 50·t - 5 = 0

From the quadratic formula, we have

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

$t = \dfrac{50\pm \sqrt{(-50)^{2}-4\times 16 \times (-5)}}{2\times 16} = \dfrac{50 \pm 2 \times \sqrt{705} }{32} = \dfrac{25}{16} \pm \dfrac{\sqrt{705} }{16}$

t ≈ 3.222 s or -0.097 s

he time the ball spends in the air ≈ 3.222 s

Therefore, Kerry's ball that spends approximately 3.222 seconds stays longer in the air than Maria's ball that spends approximately 2.64 seconds in the air.

The difference in the two time duration is 3.222 - 2.64 = 0.582

Kerry's ball stays longer in the air than Marias ball by an extra 0.582 seconds.

Answer from: Quest

i think the answer is a.( -3, -1 )

Answer from: Quest

the answer is a

step-by-step explanation: