# Math Question Help?

A stunt pilot is testing a new plane. The equation that models his height over time is h(t) = 15t^2 - 195t + 950, where t is time in seconds and h(t) is his height in meters. Determine when the stunt pilot and the test plane are below 500 meters.

Relevance

h (t) = 15t² - 195t + 950

500 >  15t² - 195t + 950

15t² - 195t + 950 - 500 < 0

15t² - 195t + 450 < 0

15 (t² - 13t + 30) < 0

15 (t - 10)(t - 3) < 0

when t < 10 and t > 3 .............ANS • h(t)=15t^2-195t+950<500

=>

t^2-13t+30<0

=>

(t-10)(t-3)<0

=>

3<t<10

=>

The plane are below 500 m

when it is more than 3 seconds

& less than 10 seconds after

starting the flight.

• We need to find the times when the pilot is at 500 metres.

i.e. when 15t² - 195t + 950 = 500

so, 15t² - 195t + 450 = 0

or, t² - 13t + 30 = 0

Hence, (t - 3)(t - 10) = 0

so, at t = 3 and again at t = 10

As the function is a parabola with a ' ∪ ' shape, the pilot is below 500 metres between these times

i.e. below 500 metres from t = 3 to t = 10 seconds

Hence, 7 seconds in all

:)>

• h(t) = 15t² - 195t + 950

We want to find time when h(t) <500

15t² - 195t + 950 < 500

15t² - 195t + 450 < 0

t² - 13t + 30 < 0

(t-10)(t-3) < 0

This gives 2 options

Option-1

(t-10)>0 and (t-3) < 0, this gives...

t>10 and t<3

This is absurd , so ignore

Option-2

(t-10)<0 and (t-3) > 0, this gives...

t<10 and t>3

That means the test plane is below 500m between 3 seconds to 10 seconds • h(t) = 15t^2 - 195t + 950  ((see red graph))

Find  15t^2 - 195t + 950 < 500, so 15t^2 - 195t + 950 - 500 < 0

Find instead where it equals 0

15t^2 - 195t + 450 = 0, solve using Quadratic

At t between 3 & 10, height will be less than 500m • set h(t) = 500 and solve for t values...then check the values of h(t) on either side of those t values