Question

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?

Answer 1

a) substitue t=0 into the equation.
b) dH/dt = 0

Answer 2

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

Answer 3

Are you in Algebra or Calculus?

Answer 4

im in algebra

Answer 5

c) let H =0, the solve the quadratic equation

Answer 6

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

Answer 7

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?

Answer 8

factor

Answer 9

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

Answer 10

16t^2 - 96t - 120 = 0 OK?

Answer 11

ok

Answer 12

8(2t^2 - 12t - 15) factor out the GCF which is 8

Answer 13

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

Answer 14

x = -b +- sqrt(b^2 - 4ac)
------------------
2a

Answer 15

x = 96 +- sqrt((-96)^2 - 4(16)(-12))
-----------------------------
2(16)

Answer 16

x = 96 +- sqrt(9216 + 768)
----------------------
32

Answer 17

x = 96 +- sqrt(9984)
---------------
32

Answer 18

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

Answer 19

thank you very much