Question

I need help with algebra!

A model rocket is launched with an initial upward velocity of 235 ft/s. The rocket's height h (in feet) after t seconds is given by the following.
h=235t-16t^2
Find all values of t for which the rocket's height is 151 feet.

Answer 1

@darkknight

Answer 2

Delete all of your comments this user needs help and STOP

Answer 3

@jhonyy9

Answer 4

@jimthompson5910

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @hhanan
Delete all of your comments this user needs help and STOP
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 6

h=235t-16t^2
151=235t-16t^2 .... replace h with 151
16t^2-235t+151 = 0 ... get everything to one side

You'll then use the quadratic formula
\(\Large t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
to find the two solutions for t.
Plug in a = 16, b = -235 and c = 151

Answer 7

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
h=235t-16t^2
151=235t-16t^2 .... replace h with 151
16t^2-235t+151 = 0 ... get everything to one side

You'll then use the quadratic formula
\(\Large t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
to find the two solutions for t.
Plug in a = 16, b = -235 and c = 151
\(\color{#0cbb34}{\text{End of Quote}}\)
i need what t=

Answer 8

If a = 16, b = -235 and c = 151
then what is \(b^2 - 4ac\) equal to?

Answer 9

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
If a = 16, b = -235 and c = 151
then what is \(b^2 - 4ac\) equal to?
\(\color{#0cbb34}{\text{End of Quote}}\)
IDK thats y im asking uuu

Answer 10

we would replace the variables with the numbers they're equal to
b^2 - 4ac = (-235)^2 - 4(16)(151) = ??

Answer 11

ok im gonna try it now

Answer 12

UGH IDK

Answer 13

What kind of calculator do you have?

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
What kind of calculator do you have?
\(\color{#0cbb34}{\text{End of Quote}}\)
ti 84

Answer 15

If you dont have one, you can use this free calculator
https://www.desmos.com/calculator

Answer 16

ok ti 84 works too

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
If you dont have one, you can use this free calculator
https://www.desmos.com/calculator
\(\color{#0cbb34}{\text{End of Quote}}\)
lol i have that to

Answer 18

whichever you prefer

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
whichever you prefer
\(\color{#0cbb34}{\text{End of Quote}}\)
ok im getting my ti

Answer 20

alright

Answer 21

i have it

Answer 22

ok type in ` (-235)^2 - 4(16)(151) `

Answer 23

i got 45561

Answer 24

same here, now apply the square root to that
\(\sqrt{45561} = \ ?\)

Answer 25

? wat

Answer 26

oohhhh i got 213.45

Answer 27

correct

Answer 28

is that my answer?

Answer 29

not yet

Answer 30

So we have this so far
\(\Large t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\Large t = \frac{-(-235)\pm\sqrt{ (-235)^2 - 4(16)(151) }}{2(16)}\)

\(\Large t = \frac{235\pm\sqrt{45561}}{32}\)

\(\Large t \approx \frac{235\pm213.45}{32}\)

\(\Large t \approx \frac{235+213.45}{32} \ \text{ or } t \approx \frac{235-213.45}{32}\)
How can we finish up?

Answer 31

t=241.67 and t=228.33

Answer 32

Not sure how you got those

Answer 33

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
Not sure how you got those
\(\color{#0cbb34}{\text{End of Quote}}\)
ugh i did 235+213.45/32

Answer 34

oh you have to add first, then divide later

Answer 35

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
oh you have to add first, then divide later
\(\color{#0cbb34}{\text{End of Quote}}\)
so what do i add?

Answer 36

The stuff up top is effectively in its own parenthesis grouping
Meaning that \(\frac{235+213.45}{32}\) is the same as \((235+213.45)/32\) and similar for the minus portion as well

So,
\(\Large t \approx \frac{235+213.45}{32} \approx \frac{448.45}{32} \approx \ ?\)

Answer 37

t=14.01 t=5226

Answer 38

I agree with t = 14.01
Not sure how you got the other value though

Answer 39

\(\color{#0cbb34}{\text{Originally Posted by}}\) @NevaehJade6219
t=14.01 t=5226
\(\color{#0cbb34}{\text{End of Quote}}\)
**15.26 i mistyped

Answer 40

What did you type in to get 15.26?

Answer 41

\(\color{#0cbb34}{\text{Originally Posted by}}\) @hhanan
@jhonyy9
\(\color{#0cbb34}{\text{End of Quote}}\)
ty you tagged me but i ve answered this problem past 1-2 hours ago

Answer 42

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
What did you type in to get 15.26?
\(\color{#0cbb34}{\text{End of Quote}}\)
488.45/32

Answer 43

how did you get 488.45?

Answer 44

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
how did you get 488.45?
\(\color{#0cbb34}{\text{End of Quote}}\)
ugh mistype again i need to redu that part it was really 448.45

Answer 45

so t=14.01and 14.01

Answer 46

The other piece you need to evaluate is \(\large t \approx \frac{235-213.45}{32}\)
You need to evaluate the stuff up top first, then divide later.

Answer 47

.67?

Answer 48

Correct, so the two approximate solutions are t = 14.01 and t = 0.67

This means that the rocket reaches the height of 151 ft at around 0.67 seconds and around 14.01 seconds

In this diagram

Point A represents when t = 0.67
Point B is when t = 14.01

Answer 49

so t=0.67 and t=14.01 is the answer?

Answer 50

Yes there are two time values when the rocket is at this height

Answer 51

oki! thanks so much old man!