Answer:
a) 112 ft.
b) 256 ft. and 3 seconds
c) 7 seconds
Step-by-step explanation:
a) The model rocket is lauched from a platform. To find the height of the platform, we need to find h when t = 0, because the rocket starts from the platform when no time has elapsed:
[tex]h=-16t^2+96t+112[/tex]
[tex]h=-16*0+96+0+112\\\\h=112[/tex]
Therefore, the height of the platform is [tex]\fbox{112}[/tex] ft.
b) If you learned calculus before, we can find the maximum height easily. We take the derivative of h and set it equal to 0. Remember, the derivative of a function is simply the slope of it at an instantaneous point. At the maximum point of a function, it's slope equals to 0.
[tex]h=-16t^2+96t+112\\h'=-32t+96+0\\h'=-32t+96[/tex]
Ok! Let's set the derivative of h to 0!
[tex]0=-32t+96\\-96=-32t\\t=3[/tex]
We now know how long it takes for the rocket to reach maximum point (t represents seconds), but we also need to find the maximum height. We can simply plug our t=3 into the function of h, because t=3 is the point where the rocket reaches maximum height:
[tex]h(3)=-16(3)^2+96*3+112\\h(3)=-144+288+112\\h(3)=256[/tex]
The maximum height of the rocket is [tex]\fbox{256}[/tex] ft and the rocket takes [tex]\fbox{3}[/tex] seconds to reach the height.
c) The rocket reaches the ground when h equals 0. We can set up the equation to solve for it:
[tex]h=-16t^2+96t+112\\0=-16t^2+96t+112\\0=-16(t+1)(t-7)\\0=(t+1)(t-7)\\t=-1, t=7[/tex]
However, time can never be negative.
Therefore, it takes the rocket [tex]\fbox{7}[/tex] seconds to reach the ground.
I hope this helps! Let me know if you have any questions :)
