Question

Calling all Algebra Smarties!! I'm stumped :/ Can anyone hep me?

Answer 1

So I'm trying to help my little brother with his homework and I am usually pretty good at it but this one is stumping me!

Answer 2

The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Answer 3

Part A: The projectile was launched from a height of 90 feet with an initial velocity of 50 feet per second. Create an equation to find the time taken by the projectile to fall on the ground.

Answer 4

well we know that v=50 and that s=90 so just substitute those values into the equation

Answer 5

Then you just need to work out at what time the height is zero, H(t) = 0

Answer 6

Okay so start with this: H(t) = -16t^2 +50t + 90

Answer 7

And then I need for figure out H and T right? Sorry haven't done this in a while :) Thanks for your help! @alekos

Answer 8

Now make H(t) =0 (height at ground) and solve for t

Answer 9

Can you help me?

Answer 10

-16t^2+50t+90=0 This is a quadratic equation which can be solved for t

Answer 11

ok heres what I got

Answer 12

t=25 ± √2065 / 16
t≈4.40264194,−1.27764194

Answer 13

@alekos

Answer 14

very good! The time cannot be -ve so the answer is 4.4 seconds to hit the ground

Answer 15

Okay so what should my final equation/answer for Part A be?

Answer 16

H(4.4) = -16(4.4)^2 +50(4.4) + 90 ??

Answer 17

hey @ayyookyndall any thoughts on how to work my problem?

Answer 18

I've already given you the equation and now you also have the solution
-16t^2+50t+90=0

Answer 19

So thats my answer? -16t^2+50t+90=0

Answer 20

Ohhh i understand now, thank you!!

Answer 21

Can you help me with the next part?

Answer 22

Part B: What is the maximum height that the projectile will reach? Show your
work.

Answer 23

I think the answer for part B is 25, is that right?

Answer 24

No, that's way out. It must be >90 because we are firing the projectile upwards at a height of 90ft

Answer 25

is the maximu 122?

Answer 26

is the maximum 129?

Answer 27

yes. its 129, but how did you get that?

Answer 28

i graphed it and found the maximum point.

Answer 29

You've really been a huge hep to me tonight. Can you help me one more time with Part C?

Answer 30

well done. but I think you may need to show an algebraic solution.
yes, go ahead with C

Answer 31

Part C: Another object moves in the air along the path of g(t) = 28 + 48.8t
where g(t) is the height, in feet, of the object from the ground at time t
seconds. Use a table to find the approximate solution to the equation H(t) =
g(t), and explain what the solution represents in the context of the problem?
[Use the function H(t) obtained in Part A, and estimate using integer values]

Answer 32

This one is really messing me up :/

Answer 33

I would just draw a graph of g(t) (straight line) and h(t) (inverted parabola) and then find where they intersect. This would mean that the two objects collide at that point.

Answer 34

You can then draw up a table of points based on the graphs if you wish

Answer 35

h(t) being what we got in part A?

Answer 36

yep

Answer 37

2.0, 125.9 is where they intercept

Answer 38

according to the graph I just made

Answer 39

yes, they collide after 2 seconds at a height of 125.9ft

Answer 40

yes!! i'm really glad i got that right!

Answer 41

okay, last one i promise! Part D?

Answer 42

ok, gotta go after D

Answer 43

Part D: Do H(t) and g(t) intersect when the projectile is going up or down,
and how do you know?

Answer 44

I think its down because when I graphed it the intersection was after h(t) had reached its maximum and started desending.

Answer 45

Max height occurs at 1.5 sec so you're right

Answer 46

see you later

Answer 47

Great thank you! Will medal and fan!

Answer 48

no problem. I enjoyed helping you out