Question

when object thrown upward direction from height h with initial velocity Vo :
using these question..v(t)=dy/dt=Vo-gt ....... y(t)=Vo-1/2gt^2
a) What is the maximum height reached by the object?
b) How long the object stays in the air?
c) With what velocity does the object hits the ground?

Answer 1

Hello Sadaf.
I want to tell you the most important thing about physics. That whenever you solve any numerical so kindly draw a pic. It solves half of your problem. Did you visualize this problem?

Answer 2

Sadaf simplifies your equations V=Vo-gt and y=Vo-0.5gt^2

Answer 3

yes i made a roughly sketch

Answer 4

okay fine....i got these eq. from following eq.
d^2 y / dt^2 = -g

Answer 5

oh My GOD so much time you take to write 4 words madam.
show your sketch. please. so we cud help you.

Answer 6

y did you differentiate them?

Answer 7

i want that you learn something from this question therefore i am not solving it for you and just helping you. kindly don't be aggressive

Answer 8

me also want to learn
i dont mind

Answer 9

i made sketch in copy

Answer 10

can you made it here?

Answer 11

sorry i mean draw it here. can you?

Answer 12

i tried but could not make

Answer 13

ok w8. let us try to do something. use simplified equations which I have given you above.

Answer 14

according to question a person standing on a height say h1. and throw a ball in upward direction. let ball goes to the height h2. so total height will be h=h1+h2 agree?

Answer 15

no ball is thrown up from ground

Answer 16

where h=0

Answer 17

"when object thrown upward direction from height h"
but these are you wordings.
ok do you know urdu language?

Answer 18

oh yes sorry this is the statement of question

Answer 19

my friend said me that ball is thrown from ground
thats y i am confused

Answer 20

yes i know urdu lang.

Answer 21

no
according to question it is from height "h" kindly check your question

Answer 22

han to madam apna question check kren. ap question smjye phle. or part(a) 2nd equation se solve hoskta hy.

Answer 23

yes u r right Sir

Answer 24

sir i checked...ball is thrown upward from height h.....not from ground sorry

Answer 25

V^2 - U^2 = 2aS right ????

Answer 26

ok then collect data.

Answer 27

a=g, S=h, so h=Vo^2/2g

Answer 28

U is initial velocity .... am i right ?

Answer 29

yes right.. now move ahead.

Answer 30

@sadaf.
Sunye. hmax=Vo^2*(sintheta)^2/2g

Answer 31

heena help her. i am not going to solve for her. kindly teach her.

Answer 32

what?? why you disagree?

Answer 33

\[h _{\max}=V _{0}^{2} \sin ^{2}(\theta)/2g\]

Answer 34

http://en.wikipedia.org/wiki/Projectile_motion

Answer 35

sorry my fault

Answer 36

ok g. ok.

Answer 37

plz shayaan help her pl

Answer 38

sir ap mujay kiyun nahe bata saktay :(

Answer 39

lol. i am asking to you and you are to me.

Answer 40

sir.. i am just 18 years old boy not sir.. okk.. dont say it again.

Answer 41

ok orry

Answer 42

acha ap data collect kren. or post kren. dekhen me ye phle hi padh chuka hn. apne ni padha hua. agr me solve kerdeta hn apk lye to ye ap apne lye apne future se cheat krengi. solve it by yourself thank you.

Answer 43

sorry

Answer 44

kindly aap solve kar dain plz mera exam hai

Answer 45

ok ok thk hy..
for (a) hmen max height find kerni hy to max height pr final velocity 0 hti hy. agree?

Answer 46

g

Answer 47

so,
Vf=0
g=9.8
and
Vi=Vo put this in your data.

Answer 48

ten mintes k bad meri light off ho jani hai so ap answer bata dena mein check kar lun ge.... or meray do swal or b hay mein post kar dun?

Answer 49

g ap post kerden. or me light meri b jane wali hy.

Answer 50

ab ap solve kerne den. w8 kren.

Answer 51

a)max height =v^2/2g
hmax=0.05Vo^2

b)v=u+at
0=Vo+gt
-Vo/g=t

Answer 52

now u know time will be same use ur mind a lil