Question

Mason launched a model rocket from the group with an initial velocity of 160 feet per second. The path of the rocket can be modeled using the equation h=-16t^2 + 160t, where h is the height above ground and t is the time after launch. At what time(s) is the rocket at a height of 336 feet?

Answer 1

Answer:

So, at time t = 3 s and t = 7 s ,Rocket will be at a height of 336 feet,

Step-by-step explanation:

Given:

initial velocity of rocket = 160 feet/s

$h=-16t^{2}+160t$

where ,h = height

            t = time

To Find:

Time = ? for h =336 feet

Solution:

$h=-16t^{2}+160t$        ......Given

Substitute h = 336 we get

$336=-16t^{2}+160t\\16t^{2}-160t+336=0$

Now Dividing the Equation throughout by 16 we get

$t^{2}-10t+21=0$

Which is a Quadratic Equation , hence on Factorizing we get

$t^{2}-7t-3t+21=0\\t(t-7)-3(t-7)=0\\(t-3)(t-7)=0\\t-3=0\ or\ t-7=0\\t=3\ or\ t=7$

Because it is a Parabolic Path ,

So, during upward journey at t = 3 s, the height will reach 336 ft.

So, it will cross the same height at another time which is at t = 7 s  

Again during downward journey, the height will be 336 from reference level at 7 s .

So, at time t = 3 s and t = 7 s ,Rocket will be at a height of 336 feet,