A ball is thrown up into the air such that its height (h) from the starting point is given by h(t) = 50t – 5t^2, where t is the time (in seconds) elapsed from the moment the ball is released. During which time interval will the height of the ball be 80 meters or higher above the starting point?

Question

anonymous · Answer

@andrewhaze @jackmullen55

anonymous · Answer

@nessachole02 @PizzaLover123

anonymous · Answer

@taylorswift131313 @ZnappyDooZ

anonymous · Answer

So what you want to do is input different values for t to find what makes the equation equal to 80. I'll do an example

anonymous · Answer

Lets say that the variable t is equal to 2

anonymous · Answer

$$80(2) = 50(2) – 5(2)^2$$

anonymous · Answer

so if I solve this equation ill get the answer?

anonymous · Answer

Simplify this and you get 
$$80 = 100 - 5(4) = 100 - 20 = 80$$
Ok well I just accidentally solved it for you

anonymous · Answer

i don't get it

anonymous · Answer

The answer is t = 2

anonymous · Answer

that's not one of my choices though that's what i don't get

anonymous · Answer

What are the choice you didn't specify

anonymous · Answer

A) ( 3,6)
B) ( 4,10)
C) (2,8)
D) ( 2,6)

anonymous · Answer

It's C

anonymous · Answer

thanks but where did u get the 8?

anonymous · Answer

I had the same question in a test once. It was (2, 8)

anonymous · Answer

never mind i got it

anonymous · Answer

Yeah