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?
Stacey Warren - Expert brainly.com
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a) substitue t=0 into the equation.
b) dH/dt = 0
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
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im in algebra
c) let H =0, the solve the quadratic equation
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
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?
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
16t^2 - 96t - 120 = 0 OK?
8(2t^2 - 12t - 15) factor out the GCF which is 8
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
x = -b +- sqrt(b^2 - 4ac)
x = 96 +- sqrt((-96)^2 - 4(16)(-12))
x = 96 +- sqrt(9216 + 768)
x = 96 +- sqrt(9984)
x = 96 +- 99.92
x = -.1225 or x = 6.1225 since x represents seconds then it would be on the
ground at 6.1 seconds