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Harlamova29_29 [7]
3 years ago
10

SEND HELPP I NEED TO PASS SCHOOL

Mathematics
1 answer:
Daniel [21]3 years ago
8 0

Answer:

I need help to pass school to

Step-by-step explanation:

so I'm going to use these point to make a question to help me pass school thank you

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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
Rama09 [41]

<em><u>Question:</u></em>

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

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

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

7 0
3 years ago
Solve for x. 2^6x+7=(1/8)^1-5x
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