Sadaf
  • Sadaf
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Shayaan_Mustafa
  • Shayaan_Mustafa
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?
Shayaan_Mustafa
  • Shayaan_Mustafa
Sadaf simplifies your equations V=Vo-gt and y=Vo-0.5gt^2
Sadaf
  • Sadaf
yes i made a roughly sketch

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Sadaf
  • Sadaf
okay fine....i got these eq. from following eq. d^2 y / dt^2 = -g
Shayaan_Mustafa
  • Shayaan_Mustafa
oh My GOD so much time you take to write 4 words madam. show your sketch. please. so we cud help you.
Shayaan_Mustafa
  • Shayaan_Mustafa
y did you differentiate them?
Shayaan_Mustafa
  • Shayaan_Mustafa
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
Sadaf
  • Sadaf
me also want to learn i dont mind
Sadaf
  • Sadaf
i made sketch in copy
Shayaan_Mustafa
  • Shayaan_Mustafa
can you made it here?
Shayaan_Mustafa
  • Shayaan_Mustafa
sorry i mean draw it here. can you?
Sadaf
  • Sadaf
i tried but could not make
Shayaan_Mustafa
  • Shayaan_Mustafa
ok w8. let us try to do something. use simplified equations which I have given you above.
Shayaan_Mustafa
  • Shayaan_Mustafa
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?
Sadaf
  • Sadaf
no ball is thrown up from ground
Sadaf
  • Sadaf
where h=0
Shayaan_Mustafa
  • Shayaan_Mustafa
"when object thrown upward direction from height h" but these are you wordings. ok do you know urdu language?
Sadaf
  • Sadaf
oh yes sorry this is the statement of question
Sadaf
  • Sadaf
my friend said me that ball is thrown from ground thats y i am confused
Sadaf
  • Sadaf
yes i know urdu lang.
Shayaan_Mustafa
  • Shayaan_Mustafa
no according to question it is from height "h" kindly check your question
Shayaan_Mustafa
  • Shayaan_Mustafa
han to madam apna question check kren. ap question smjye phle. or part(a) 2nd equation se solve hoskta hy.
Sadaf
  • Sadaf
yes u r right Sir
Sadaf
  • Sadaf
sir i checked...ball is thrown upward from height h.....not from ground sorry
Sadaf
  • Sadaf
V^2 - U^2 = 2aS right ????
Shayaan_Mustafa
  • Shayaan_Mustafa
ok then collect data.
Sadaf
  • Sadaf
a=g, S=h, so h=Vo^2/2g
Sadaf
  • Sadaf
U is initial velocity .... am i right ?
Shayaan_Mustafa
  • Shayaan_Mustafa
yes right.. now move ahead.
Shayaan_Mustafa
  • Shayaan_Mustafa
@sadaf. Sunye. hmax=Vo^2*(sintheta)^2/2g
Shayaan_Mustafa
  • Shayaan_Mustafa
heena help her. i am not going to solve for her. kindly teach her.
Shayaan_Mustafa
  • Shayaan_Mustafa
what?? why you disagree?
Shayaan_Mustafa
  • Shayaan_Mustafa
\[h _{\max}=V _{0}^{2} \sin ^{2}(\theta)/2g\]
Shayaan_Mustafa
  • Shayaan_Mustafa
http://en.wikipedia.org/wiki/Projectile_motion
anonymous
  • anonymous
sorry my fault
Shayaan_Mustafa
  • Shayaan_Mustafa
ok g. ok.
anonymous
  • anonymous
plz shayaan help her pl
Sadaf
  • Sadaf
sir ap mujay kiyun nahe bata saktay :(
Shayaan_Mustafa
  • Shayaan_Mustafa
lol. i am asking to you and you are to me.
Shayaan_Mustafa
  • Shayaan_Mustafa
sir.. i am just 18 years old boy not sir.. okk.. dont say it again.
Sadaf
  • Sadaf
ok orry
Shayaan_Mustafa
  • Shayaan_Mustafa
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.
Sadaf
  • Sadaf
sorry
Sadaf
  • Sadaf
kindly aap solve kar dain plz mera exam hai
Shayaan_Mustafa
  • Shayaan_Mustafa
ok ok thk hy.. for (a) hmen max height find kerni hy to max height pr final velocity 0 hti hy. agree?
Sadaf
  • Sadaf
g
Shayaan_Mustafa
  • Shayaan_Mustafa
so, Vf=0 g=9.8 and Vi=Vo put this in your data.
Sadaf
  • Sadaf
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?
Shayaan_Mustafa
  • Shayaan_Mustafa
g ap post kerden. or me light meri b jane wali hy.
Shayaan_Mustafa
  • Shayaan_Mustafa
ab ap solve kerne den. w8 kren.
anonymous
  • anonymous
a)max height =v^2/2g hmax=0.05Vo^2 b)v=u+at 0=Vo+gt -Vo/g=t
anonymous
  • anonymous
|dw:1332521837319:dw| now u know time will be same use ur mind a lil
anonymous
  • anonymous
u knw time will be same so equate dem v1-u1=at......................................(1) here v1=0 , u1=Vo V2-U2=at.....................................(2) here U2=0, V2=? V1-U1-U2=-V2
anonymous
  • anonymous
any doubt?
anonymous
  • anonymous
is this necessarry to do so as there are so many ways.. (: but like to see how can we use this equation u have any idea @eashmore
anonymous
  • anonymous
hey i get this equation now v(t)=dy/dt=Vo-gt dy/dt=V here V =final velocity which is equal to v(t) now we know V=u +gt and u=initial velocity which is Vo but my dout how -ve sign comes :?
anonymous
  • anonymous
This is a rather simple problem. Note that at the top of the trajectory, \(v =0\) Therefore, \[0 = v_o - g t\]We can solve this equation for t, yielding \[t = {v_o \over g}\] To find the height reached at this time, simply plug this expression for time into our equation for height\[y = v_0t - {1 \over 2} g t^2 = {v_0^2 \over g} - {1 \over 2} {v_0^2 \over g} = {v_0^2 \over 2g}\] The time spent in flight is simply\[t = {2v_0 \over g}\] The object will have the same velocity when it hits the ground as it did at launch.
anonymous
  • anonymous
i agree but i need a solution by using the equation given in the qn
anonymous
  • anonymous
I used the equation presented in the question.

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