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A toy rocket is launched vertically upward from a 8 foot platform with an
initial velocity of 120 feet per second. Its height h at time t seconds after
launch is given by the equation h(t) = -16t^2 + 120t + 8. How many seconds
until the rocket is 40 feet? (There may be more than one answer)
A) 0.28
B) 0.94
C) 3.17
D) 4.62
E) 5.98
F) 6.44
G) 7.22
H) 8.17

Answer:

We get t= 7.22 and t=0.28

The correct options are Option A and Option G

Step-by-step explanation:

The height h of toy rocket at time t seconds after  launch is given by the equation [tex]h(t) = -16t^2 + 120t + 8[/tex]

We need to find how many seconds  until the rocket is 40 feet?

We are given height of rocket h(t) = 40 feet

and we need to find time t

Using the equation [tex]h(t) = -16t^2 + 120t + 8[/tex] we can find time t

We have h(t)=40

[tex]h(t) = -16t^2 + 120t + 8\\Put\:h(t)=40\\40= -16t^2 + 120t + 8\\16t^2-120t-8+40=0\\16t^2-120t+32=0[/tex]

Now, we will solve the quadratic equation using quadratic formula: [tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

We have a =16, b=-120 and c=32

Putting values and finding t

[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\t=\frac{-(-120)\pm\sqrt{(-120)^2-4(16)(32)}}{2(16)}\\t=\frac{120\pm\sqrt{14400-2048}}{32}\\t=\frac{120\pm\sqrt{12352}}{32}\\t=\frac{120\pm111.12}{32}\\t=\frac{120+111.12}{32}, t=\frac{120-111.12}{32}\\t=7.22\:,\:t=0.28[/tex]

So, we get t= 7.22 and t=0.28

The correct options are Option A and Option G

PLZZ HELPP IM FAILLING THIS CLASS

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