anonymous
  • anonymous
a rocket is fired into the air from the top a building. The equation H= -16t2 + 96t + 120 gives H, the height of the rocket in feet, t seconds after it was fired. a) How tall was the building? b) What was the maximum height of the rocket? c) When did the rocket hit the ground?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
a) substitue t=0 into the equation. b) dH/dt = 0
anonymous
  • anonymous
a. well when time = 0 the rocket has not fired yet and it is sitting on top of the building so H = -16(0) + 96(0) + 120
anonymous
  • anonymous
Are you in Algebra or Calculus?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
im in algebra
anonymous
  • anonymous
c) let H =0, the solve the quadratic equation
anonymous
  • anonymous
OK. so we need to find the maximum of the parabola. It is found when x = -b/2a (x in this case is t) so t = -96/-32 which is t = 3 that means when t = 3 or after 3 seconds, it is at its highest point. to find out how high that point is put in 3 for t H = -16(3^2) + 96(3) + 120
anonymous
  • anonymous
To find out when it hits the ground, we need to find out when H = 0 0 = -16t^2 + 96t + 120 we need to factor this or use the quadratic formula. which do you want to do?
anonymous
  • anonymous
factor
anonymous
  • anonymous
OK.. you always want the t^2 to be positive so we are going to take it over to the other side of the equation
anonymous
  • anonymous
16t^2 - 96t - 120 = 0 OK?
anonymous
  • anonymous
ok
anonymous
  • anonymous
8(2t^2 - 12t - 15) factor out the GCF which is 8
anonymous
  • anonymous
Well... it doesn't factor so we have to use the quadratic formula. I am going back to the original 16t^2 - 96t - 120 = 0
anonymous
  • anonymous
x = -b +- sqrt(b^2 - 4ac) ------------------ 2a
anonymous
  • anonymous
x = 96 +- sqrt((-96)^2 - 4(16)(-12)) ----------------------------- 2(16)
anonymous
  • anonymous
x = 96 +- sqrt(9216 + 768) ---------------------- 32
anonymous
  • anonymous
x = 96 +- sqrt(9984) --------------- 32
anonymous
  • anonymous
x = 96 +- 99.92 ------------ 32 x = -.1225 or x = 6.1225 since x represents seconds then it would be on the ground at 6.1 seconds
anonymous
  • anonymous
thank you very much

Looking for something else?

Not the answer you are looking for? Search for more explanations.