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23 Feb 2007 18:52:43 IST
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Asmita Bhattacharya
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23 Feb 2007 19:05:31 IST
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We've to find the time t when v=10 m/s
Since at t=0 v= 20m/s at an < of 60 0 with the vertical vx=10rt3 m/s ,vy=10 m/s
So v=10m/s when vy=0
0=10-gt
so t=1 second
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23 Feb 2007 19:26:51 IST
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initial speed = 20 m/s.horizontal component of this velocity = 10 m/s. final asked speed = 10m/s. this means that when speed is 10 m/s it is at max. height as vertical component of velocity =0. therfore t =1/2(time period of progectile) = 1/2 (2 u sin theta/g)=u sin theta /g = 
3 seconds.
3 seconds.
3 m/s
23 Feb 2007 21:35:59 IST
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An easy way:
apply conservation of energy.....initial velocity = v and assume GPE = 0 at projecting posistion.....velocity is v/2 whn hight is h....
1/2 mv^2 = mgh + 1/2 m(v/2)^2
putting the values h = 15 meter.....
initial velocity in y direction is Vy = v cos60 (or v sin30 since angle with horizontal is 30)........===>> Vy = 20 cos 60 = 10 m/s
h = ut + 1/2 gt^2
put H = 15 , u (Vy)= 10 , g=10
Solving you get t = 1 sec
apply conservation of energy.....initial velocity = v and assume GPE = 0 at projecting posistion.....velocity is v/2 whn hight is h....
1/2 mv^2 = mgh + 1/2 m(v/2)^2
putting the values h = 15 meter.....
initial velocity in y direction is Vy = v cos60 (or v sin30 since angle with horizontal is 30)........===>> Vy = 20 cos 60 = 10 m/s
h = ut + 1/2 gt^2
put H = 15 , u (Vy)= 10 , g=10
Solving you get t = 1 sec
24 Feb 2007 01:14:50 IST
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@ ashmita
I agree with ur answer in ur second post....nd thr is no mistakr conceptually....thr might b a bug in the question....evn on solving my part we get t = 1 nd -3....nd -3 is irrelevent seeing the question...
In ne case ur concept is correct
I agree with ur answer in ur second post....nd thr is no mistakr conceptually....thr might b a bug in the question....evn on solving my part we get t = 1 nd -3....nd -3 is irrelevent seeing the question...
In ne case ur concept is correct
26 Feb 2007 09:59:25 IST
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Its initial horizontal velocity is 20cos600 = 10 m/s
Initailly total speed = 20 m/s
When the projectile becomes horizontal (its vertical velocity becomes zero), the projectile is at maximum height and its speed at this point is half of its initial speed.
Vertical velocity = 20sin600 - gt = 0
t = 20sin600/10 =
3 sec.
Best Wishes
Initailly total speed = 20 m/s
When the projectile becomes horizontal (its vertical velocity becomes zero), the projectile is at maximum height and its speed at this point is half of its initial speed.
Vertical velocity = 20sin600 - gt = 0
t = 20sin600/10 =
3 sec.Best Wishes
26 Feb 2007 17:12:25 IST
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@ iitkgp bipin.....plz read the question again it says 60 with vertical and not with horizontal and chk ur solution again.....
@mandar5227 .....Dude jst chk the question and solve it urself b4 posting such words "ALL ARE WRONG ONLY KGHEDRIU IS CORRECT. GOOD DUDE" for neone else.....
@mandar5227 .....Dude jst chk the question and solve it urself b4 posting such words "ALL ARE WRONG ONLY KGHEDRIU IS CORRECT. GOOD DUDE" for neone else.....
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