Math question *grade 10*

Question

. A model rocket is shot straight up from the roof of a school. The height in metres at any time t, in seconds, is given by h = − 5t 2 +40t. How long does is the rocket above 60 metres in height?

Answer 1

-5t^2 + 40t = 60

t^ -8t + 12 = 0

t= 8/2 + or - (1/2)sqr(64-48) = 4 + or - sqr(16)/2

= 6 or 2 seconds

-5(2)^2 + 40(2) = -20+80 = 60

-5(6)^2 + 40(6) = -180+240 = 60

on the way up, it reaches 60 meters after 2 seconds

then on the way down it again reaches 60 meters after 6 seconds

h(t) = -5t^2 + 40t

h'(t) = -10t +40 = 0

t = 40/10 = 4 seconds to reach maximum height

h(4) = -5(4)^2 + 40(4) =-5(16) + 160 = 160-80= 80 feet max

2 seconds before and 2 seconds after reaching max height of 80 feet, it reaches 60 feet at 4-2=2 and 4+2= 6 seconds

initial height = 0

initial velocity = 40 feet per second

effect of gravity = 10 feet per second per second deceleration

(but actually, it's 9.8 ft/sec^2, but this problem rounded it off to -10. -10/2= -5 is the coefficient of the t^2 term. at^2/2 where a really = -9.8 and a/2 = 9.8/2 = 4.9)