Mathematics, 25.02.2020 09:23 hhh4775

Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground? 1 s 2 s 3 s 4 s

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Answer from: halllawson

It will take 2 seconds for the object to hit the ground.

Given that,

Britney throws an object straight up into the air with an initial velocity of 27 feet/sec.

From a platform that is 10 feet above the ground.

Formula; h(t)= -16t^2+v_ot+h_0 ,

Where v_0 is the initial velocity and h_0 is the initial height.

We have to determine,

How long will it take for the object to hit the ground.

According to the question,

Formula; h(t)= -16t^2+v_ot+h_0

Where,

$h_0 = 10 \ feet \ and \\ v_0 = 27\ feet \ per \ second$

Substitute the values in the given formula,

$h(t)= -16t^2+v_ot+h_0\\\\h(t) = -16t^2 + 27t + 10$

When the point the object hits the ground, h(t) become 0,

Then,

$\h(t) = -16t^2 + 27t + 10\\\\0 = -16t^2 + 27t + 10\\\\ -16t^2 + 27t + 10 = 0$

For the quadratic equation solution is determined by the formula.

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

Where, a = -16, b = 27, c = 10

Substitute all the values in the formula;

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x = \dfrac{-(-27)\pm\sqrt{(-27)^2-4(16).(10)}}{2(16)}\\\\\\x = \dfrac{27\pm\sqrt{729+640}}{32}\\\\x = \dfrac{27\pm\sqrt{1369}}{32}}\\\\x = \dfarc{27\pm37}{32}\\\\x = \dfrac{27+37}{32} \ and \ x = \dfrac{27-37}{32}\\\\x = \dfrac{64}{32} \ and \ x= \dfrac{-10}{32}\\\\x = 2 \ and \ x = - 0.3125$

Time can not be negative. so, x = -0.3125 rejected.

Hence, It will take 2 seconds for the object to hit the ground.

To know more about Quadratic equation click the link given below.

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Answer from: morganruhl2

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

Answer from: Quest

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)

Answer from: Quest

0.038

step-by-step explanation:

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Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. H...

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