Question

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle?

128 feet
224 feet
272 feet
484 feet

Answer 1

First thing I did was get the formula for the gravity.

s(t) = -16t^2 + 144t + 160

I plugged it in. What do I do next?

Answer 2

@Jadeishere

Answer 3

The formula would be
s(t) = -16t^2 +vt + s
Plug in the values;
s(t) = -16t^2 + 144t + 160

Answer 4

What next?

Answer 5

I'm not quite sure.. Like I said, not the best with math. Anyone would be better than me.
@agent0smith @AloneS @imqwerty @SIenderman

Answer 6

@welshfella

Answer 7

@sammixboo

Answer 8

find the derivative ds /dt This will give you an expression for the velocity. When the projectile is at its maximum height the value of v will be 0.

v = ds/dt = -32t + 144 = 0
t = 144/32 = 4.5 seconds

Answer 9

Now to find maximum height
s(4.5) = -16(4.5)^2 + 144(4.5) + 160

Answer 10

grab ur calculator and work this out

Answer 11

have you done any calculus? derivatives?

Answer 12

No

Answer 13

If not then we can use another constant acceleration question ( which came form calculus anyway)

v = u + at

u = initial speed = 144, a = acceleration due to gravity = -32 , and v = 0 9 at maximum height)

0 = 144 - 32t

- giving t = 4.5 secs

Answer 14

I just dont understand how to get t?

Answer 15

@welshfella

Answer 16

0 = 144 - 32t

solve for t

subtract 144 from both sides
-144 = -32t

now divide both sides by -32

Answer 17

Where is 32 from? Is that a second problem?

Answer 18

to solve for t its 0 = v - st ?

Answer 19

You need help?

Answer 20

Yes @FaiqRaees

Answer 21

First I am going to clear the assumptions we're taking in this question(Since the question is vague in every aspect)
1. The object has initial velocity in the vertical direction
2. Gravitational acceleration is opposite to the initial velocity
3. da/dh = 0 and acceleration =9.81
Clear?

Answer 22

My tacher said do -b/2a.... like aos

Answer 23

We're going to suppose the initial height is 0. (Meaning our ground has risen to 160 feet)
Use the formula
\[\large\rm (V_{final})^2=(V_{initial})^2-2(acceleration)(distance)\ \]
\[ \color{red}{\text{ Note:- The sign after initial is changed because acceleration is negative.}} \]

Answer 24

When I do that I get 4.5.
s(4.5) = -16(4.5)^2 + 144(4.5) + 160
When I do that I get 252...

Answer 25

What values are you taking for each variable?

Answer 26

t = 4.5
a: -16
b: 144

Answer 27

-144/-32 is 4.5.

Answer 28

Which formula are you using?

Answer 29

-b/2a

Answer 30

That's a wrong formula, use the one I provided

Answer 31

WHen I am done I get 484. Is that the answer?

Answer 32

Correct

Answer 33

What I did was WAY easier.. XD
I did simply -b/2a and got t. Then I solved for s(t) and got it :P thank you!

Answer 34

Good

Answer 35

But in my method you just had to plug the respective values and then add 160

Answer 36

I havent learned that yet.. I dont even know what a respective value it. This is Algebra 1 and im 13 XD

Answer 37

Well then you should next time specify that. In this way the others will keep the proficiency level of their explanation to minimal

Answer 38

Sorry D:

Answer 39

lol poor guy first got calculus help, then physics help. Probably a good idea to next time specify that you're in algebra 1 @OswaldMurphy since these type of questions don't make it clear what type of class you're in.

Answer 40

Super sorry :( I was busy doing other stuff and I should've stayed here to help you... @OswaldMurphy

Answer 41

@agent0smith I didnt know this question had different solutions. Thanks @Jadeishere

Answer 42

Its ok @Jadeishere *

Answer 43

Well it can be done as a physics problem, an algebra problem, or a calculus problem. They'll all get the same answer, but they're all pretty different methods.