Model Rocket Launched from a Hill overlooking a lake

Question

A model rocket is launched from a hill overlooking a lake. The height of the rocket, measured in feet, after t seconds is given by the function.

h(t)=−40t^ 2+245t+30

A). At what time(s) is the height of the rocket equal to 300 feet? Give your answer(s) rounded to three decimal places

B). How long is the rocket in the previous problem in the air?

Answer 1

-Estimated answers are 1.5 seconds, and 4.5 seconds for 300 feet, and then double the midpoint of 3 seconds to get about 6 seconds for hitting the lake

40t²+245t+30 =300

divide by 5

-8t² +49t + 6= 60

-8t² + 49t -54 = 0 use the quadratic formula to solve for t a=-8, b= 49 c=-54

t = [-49 + or - sqr(49² -4(-8)(-54)]/-16

= -49/16 + or - 1/16 sqr (2401-1728)

= -49/16 + or - 1/16 sqr (673)

= 49/16 + or - 1/16 (25.942)

= 23.058/16 or 73.942/16 seconds

=1.438 seconds or 4.621 to be at 300 feet

almost 1 1/2 seconds going up, then coming down to 300 feet a little over 4 1/2 seconds. Right in the middle, at t=3, the rocket reaches maximum height. Double that, and about t=6 is when it hits the lake, approximately.

to find when h=0 set the original equation = 0 and solve for t

5( -8t² + 49t +6) = 0 apply the quadratic formula a=-8, b=49 c=6

t=[-49 + or - sqr(49² - 4(-8)(6)]/-16

= 49/16 + or - 1/16 sqr(2401 + 192)

=(49 + or - 50.922)/16

= 99.922/16 or -1.922/16 seconds = 6.245 seconds before it reaches h=0, ignore negative solution

plug 6 into -8t²+49t+6 = -8(36)+49(6)+6 = -288+295+6=13 6 seconds is still a little above h=0 in the next quarter second it goes down another 13 feet

the original equation translates into -40t² as acceleration downward due to gravity, 245t = initial velocity, 245 feet per second, and 30 feet as initial height above sea level, -16t² would be the usual gravity deceleration. -40t² would seem to be a mistake or maybe just a different "gravity" for a denser planet than earth

take the first derivative -80t + 245 = 0 and solve for t to get the maximum height t=245/80=49/16=3.0265 seconds, substitute that into the original equation to get maximum height h(49/16)=40(49/16)²+245(49/16)+30

= 40(2401/256)+192080/256+30=(96040+192080)/256 + 30 =288120/256 + 30 =1125.469 feet maximum after slightly over 3 seconds

the time to reach maximum height is about 1/2 the time to reach h=0

It would be exactly half had the rocket been launched from sea level instead of 30 feet above.