Mathematics, 03.12.2019 23:31 76022sfox

Use the position equation s = −16t2 + v0t + s0 where s represents the height of an object (in feet), v0 represents the initial velocity of the object (in feet per second), s0 represents the initial height of the object (in feet), and t represents the time (in seconds). a projectile is fired straight upward from ground level (s0 = 0) with an initial velocity of 128 feet per second.(a) at what instant will it be back at ground level? (b) when will the height be less than 128 feet? (enter your answer using interval notation.) i need with part b. it is supposed to be an interval notation and look like this u ex.[0,6-2 radical6) u (6+2radical6,12)

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Answer from: CameronVand21

After 8 seconds, the object will be back at ground level.

At 1.2 seconds, the object will be in a position less than 128 feet.

Given that,

The position equation s = $-16t^{2} +v_o+s_0$

where s represents the height of an object (in feet),

v_0 represents the initial velocity of the object (in feet per second),

s_0 represents the initial height of the object (in feet),

and t represents the time (in seconds).

initial velocity of 128 feet per second.

According to the question,

Using the equation given, we have to derive, because the derivative of a displacement functions results in velocity, and then we just calculate time, like this:

$s = -16t^{2} +v_o+s_0$

Where $s_0 = 0 , v_0 = 128\frac{ft}{sec}$

Putting the given values,

$s_0 = -16t^{2} +128t +0$

Differentiate both the sides,

$v_0 = 2(-16)t^{2-1} +128\\v_0 = -32t +128$

When the object reach the highest point, the velocity in that moment is zero. (v = 0)

So, -32t + 128 = 0

32t = 128

t = $\frac{128}{32}$

t = 4 sec

Therefore, the object takes 4 seconds long to reach the highest point. Also, this is a symmetrical movement (because is a quadratic function), this means that the object uses double of this time to reach the ground. So, after 8 seconds, the object will be back at ground level.

To solve the second part of the problem, we have to solve a quadratic inequality. A height less than 128 feet can be expressed like:

$s = -16t^{2} + 128$

Replacing the inequality

$-16t^{2} + 128t < 128\\-16t^{2} + 128t - 128 < 0$

Divided by 16 both the sides,

$x^{2} -8t+8 0$

Therefore, solving this quadratic equation, the roots are = 6.82843 and 1.17157. So, the answer that makes sense is = 1.17157, because the total time to reach the highest point is 4 and 6 seconds is more than 4.

Hence, at 1.2 seconds, the object will be in a position less than 128 feet.

For the more information about the Average Velocity click the link given below.

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Answer from: paulinahunl17

Using the equation given, we have to derive, because the derivative of a displacement functions results in velocity, and then we just calculate time, like this:

$s=-16t^{2}+v_{0}t+s_{0}$

We know that $s_{o}=0$ and $v_{0}=128ft/sec$

Replacing these values: $s=-16t^{2} +128t+0$

Then, we derive: $v=2(-16)t^{2-1} +128x^{1-1}\\ v=-32t+128t^{0} \\v=-32t+128$

But, when the object reach the highest point, the velocity in that moment is zero. (v = 0)

So, $0=-32t+128\\-128=-32t\\\frac{-128}{-32} =t\\t=4sec$ .

To solve the second part of the problem, we have to solve a quadratic inequality. A height less than 128 feet can be expressed like:

We know that $s=-16t^{2} +128t$

Replacing this s expression on the inequality:

$-16t^{2}+128t$

Solving this inequality:

$-16t^{2} +128t-128$

Dividing by -16:

$\frac{-16t^{2}+128t-128 }{-16}0$ ; which is easier too solve.

Therefore, solving this quadratic equation, the roots are = 6.82843 and

=1.17157. So, the answer that makes sense is =1.17157, because the total time to reach the highest point is 4 and 6 seconds is more than 4.

Hence, at 1.2 seconds, the object will be in a position less than 128 feet.

Answer from: Quest

$35.59 in pounds

step-by-step explanation:

Answer from: Quest

only 10/60 isn't equivalent

hope this : )

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