Question

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 HELP 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)

Answer 1

Answer:

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

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: $s$

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.