A rocket is fired upward with an initial velocity v of 100 meters per second. The quadratic function S(t)=-5t^2+100t can be used to find the height s of the rocket, in meters, at any time t in seconds. Find the height of the rocket 7 seconds after it takes off. During the course of its flight, after how many seconds will the rocket be at a height of 450 meters?

The rocket will be at the height of 450metres at 13.16secs and 6.84secs

Given the expression modeled by the height S(t)=-5t^2+100t where

t is the time taken by the rocket to take off

s is the height traveled by rocket

In order to find the height of the rocket 7 seconds after it takes off, we will substitute t = 7 into the equation

S(7) = -5(7)²+100(7)

S(7) = -5(49)+700

S(7) = -245+700

S(7) = 455metres

Hence the height of the rocket 7 seconds after it takes off is 455metres

Given that S = 450m, we can also get the time taken by the rocket at this height.

Recall that S(t)= -5t²+100t

450 = -5t²+100t

Rearrange

-5t²+100t - 450 = 0

5t²-100t + 450 = 0

Divide through by 5

t²-20t + 90 = 0

On factorizing above equation;

t= 10+√10 or t=10−√10

t = 10+3.1623 or 10 - 3.1623

t = 13.16 and 6.84secs

Hence the rocket will be at the height of 450metres at 13.16secs and 6.84secs

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b

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