Question

Hi, Can someone please help me show that the max height attained by a ball projected in air can me model by h~1.2t^2 where t is time in seconds that ball is in air. For example ball was in air 5 sec and max height was 30 m. THe problem is asking me to prove that this formula can be true...

Answer 1

h = 1.2t^2?

Answer 2

yes, it is asking me to show that thois can model max height

Answer 3

is this college physic? like 2000 level?

Answer 4

PH 211. Im having alot of trouble even with the simplest things because we dont do any examples of any kind in lectures );

Answer 5

I'm not sure what kind of answer is wanted, but if you take the derivative of the h = 1.2t^2 equation, it will give you the 'maximum height' ..

Answer 6

I have a webpage open with the same problem if you would like to see it but I still am confused by it.

Answer 7

a class with few examples, yuck!

Answer 8

Yeah I like the stuff once I get it but hours of surfing the net and youtube after class is tiring.

Answer 9

only teacher and course offered at my school for my program lol

Answer 10

Read :: http://www.fullmarks.org/images/stories/studytips/physics-xi-xii.pdf

Answer 11

ususally max height is a function of initail velocity in the y direction.... but like if we were to graph what this looks like the graph would look like this:

Answer 12

http://answers.yahoo.com/question/index?qid=20091008021233AAg8ZUD Im trying to make sense of this which is a proof.

Answer 13

ohh, ok ok, i get it, one sec

Answer 14

Thank you.

Answer 15

so do you know the equation s = s + vt + (1/2)at^2 ?

Answer 16

position eq, yeah?

Answer 17

I know max height should be at t/2....

Answer 18

Can I plug that into position formula?

Answer 19

yikes, this is taking me a lot longer than I thought, almost there tho!

Answer 20

lol thanks again

Answer 21

I think I got it Wolf

Answer 22

h(t)=Xo+Vo+1/2at^2 and plug in t/2 spits out h=1.2t^2 once you put in gravitational constant

Answer 23

I'm still puzzled by this, the yahoo answers doesn't make sense to me.. but i'm not a physics teacher, just have taken the class!

Answer 24

shoot, I can't figure it, sorry man. I keep getting h=4.9t^2

Answer 25

ohh I got it I got it!

if you still need help I can help you!
I wassn't thinking of t as total time! I got it though if you wnt me to explain it :D

Answer 26

Vf = Vi + at, [kinematic equation]
0 = Vi + at
Vi = -at

S = S + vt + (1/2)at^2 [kinemaitc equation]
h = 0 +(-at)t + (1/2)at^2
h = -at^2 + .5at^2
h = -.5at^2
h = -.5*-9.8*t^2
h = 4.9t^2

where total time, T, equals 2t, so T=2t, or t=T/2
h = 4.9 (T/2)^2 becomes
h = 4.9 (T^2/2^2) becomes
h = 4.9 T^2 (1/4) becomes
h = 1.2 T^2

boo ya, lol

Answer 27

Thanks alot wolf!

Answer 28

do you have any questions about what I did at any point?
oh, btw, the reason, on the second kinematic eqiation, v = 0 is becuase in the y direction, when the projectile reaches it's peak, it has no more speed going up, and for an instant is equal to zero, then gravity kicks in and starts pulling it back down.