Answered

IDNLearn.com: Your destination for reliable and timely answers to any question. Our platform provides trustworthy answers to help you make informed decisions quickly and easily.

An object projected upwards with a velocity of 96 feet per second from a height of 6 feet above theground is modelled by the function ℎ() = −162 + 96 + 6 .A. [3 pts] How many seconds after launch will the object reach its maximum height? Round your answerto one decimal place.B. [3 pts] Find the maximum height that the object reaches. Round your answer to one decimal place.C. [3 pts] Find the x-intercept and explain its meaning on the context of the problem.D. [3 pts] After how many seconds will the object be 100 feet above the ground?E. [2 pts] Find the y-intercept and explain its meaning on the context of the problem

Sagot :

Given:

[tex]h(t)=-16t^2+96t+6[/tex]

Find-:

(a) Maximum second after launch will the object reach its maximum height

(b) Find the maximum height that the object reaches.

(c) Find the x-intercept and explain its meaning in the context of the problem.

(d) After how many seconds will the object be 100 feet above the ground

(e) Find the y-intercept and explain its meaning on the context of the problem

Sol:

(a)

Maximum second after launch.

For maximum value derivative should be zero.

[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h^{\prime}(t)=-(16\times2)t+96 \\ \\ \end{gathered}[/tex]

[tex]\begin{gathered} -32t+96=0 \\ \\ 32t=96 \\ \\ t=\frac{96}{32} \\ \\ t=3 \end{gathered}[/tex]

After 3-second the object reaches maximum height.

(b)

For maximum height is at t = 3

[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(3)=-16(3)^2+96(3)+6 \\ \\ h(3)=(-16\times9)+(96\times3)+6 \\ \\ h(3)=-144+288+6 \\ \\ =150 \end{gathered}[/tex]

(c) x-intercept the value of y is zero that means:

[tex]\begin{gathered} h(t)=0 \\ \\ -16t^2+96t+6=0 \\ \\ -8t^2+48t+3=0 \\ \\ t=\frac{-48\pm\sqrt{48^2-4(-8)(3)}}{2(-8)} \\ \\ t=\frac{-48\pm48.98}{-16} \\ \\ t=6;t=-0.061 \end{gathered}[/tex]

The negative value of "t" is not considered so at

x-intercept is 6 and -0.061

(d) Object be 100 feet above grounded is:

[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ -16t^2+96t+6=100 \\ \\ -16t^2+96t-94=0 \\ \\ -8t^2+48t-47=0 \\ \end{gathered}[/tex]

So, the time is:

[tex]\begin{gathered} t=\frac{-48\pm\sqrt{48^2-4(-8)(-47)}}{2(-8)} \\ \\ t=\frac{-48\pm\sqrt{800}}{-16} \\ \\ t=\frac{-48-28.28}{-16},t=\frac{-48+28.28}{-16} \\ \\ t=4.76,t=1.23 \end{gathered}[/tex]

At t= 4.76 and t =1.23

(e)

For y-intercept value of "x" is zero.

[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(0)=-16(0)^2+96(0)+6 \\ \\ h(0)=6 \end{gathered}[/tex]

So, the y-intercept is 6.

Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.