Question

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

Answer 1

Given:

`h(t)=-16t^2+96t+6`

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.

`\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h^(\prime)(t)=-(16*2)t+96 \\ \\ \end{gathered}`

`\begin{gathered} -32t+96=0 \\ \\ 32t=96 \\ \\ t=(96)/(32) \\ \\ t=3 \end{gathered}`

After 3-second the object reaches maximum height.

(b)

For maximum height is at t = 3

`\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(3)=-16(3)^2+96(3)+6 \\ \\ h(3)=(-16*9)+(96*3)+6 \\ \\ h(3)=-144+288+6 \\ \\ =150 \end{gathered}`

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

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

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:

`\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}`

So, the time is:

`\begin{gathered} t=(-48\pm√(48^2-4(-8)(-47)))/(2(-8)) \\ \\ t=(-48\pm√(800))/(-16) \\ \\ t=(-48-28.28)/(-16),t=(-48+28.28)/(-16) \\ \\ t=4.76,t=1.23 \end{gathered}`

At t= 4.76 and t =1.23

(e)

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

`\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(0)=-16(0)^2+96(0)+6 \\ \\ h(0)=6 \end{gathered}`

So, the y-intercept is 6.