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3 years ago

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If a baseball player hits a baseball from 4 feet off the ground with an initial velocity of 64 feet per second, how long will it

take the baseball to hit the ground? Use the equation h = –16t2 + 64t + 4.
A. 2 plus or minus square root of 17 end root over 2

B. quantity of 2 plus or minus square root of 17 all over 2

C. 2 plus or minus 4 square root of 17

D. quantity of 16 plus or minus square root of 17 all over 2

2 answers:

hichkok12 [17]3 years ago

8 0

<h3><u>Answer;</u></h3>

A. <em><u> $x = 2 +/- sqrt(17) / 2$ </u></em>

<h3><u>Explanation;</u></h3>

Using the equation given we can calculate the value of t, time taken by the basket ball to hit the ground.

$h = -16t^2 + 64t + 4 \\0 = -16t^2 + 64t + 4 \\16t^2 - 64t - 4 = 0 \\4(4t^2 - 16t - 1) = 0 \\4t^2 - 16t - 1 = 0 \\t = (-b +/- sqrt(b^2 - 4ac)) / 2a \\t = (16 +/- sqrt(256 + 16)) / 8 \\t = (16 +/- sqrt(272)) / 8 \\t = (16 +/- 4 sqrt(17)) / 8 \\t= 2 +/- sqrt(17) / 2$

Tpy6a [65]3 years ago

5 0

Answer:

$t = 2 \pm \frac{\sqrt{17}}{2}$

Explanation:

As we know that the position of the ball is related with time given as

$h = -16 t^2 + 64 t + 4$

now when ball hit the ground then we have

h = 0

so we will have

$-16 t^2 + 64 t + 4 = 0$

$-4 t^2 + 16 t + 1 = 0$

so here by solving the above equation we have

$t = \frac{-16 \pm \sqrt{256 + 16}}{-8}$

$t = 2 \pm \frac{\sqrt{17}}{2}$

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