11theta is tan^-1(2/5)
time period is 20/sqrt(29)
This is a very easy question in projectile...
Everyone can solve it so please solve and give if not the solution but at least the outline of your method...
See finally which way it goes..
Q. Air provides 4ms-2 acceleration to right. In this situation a boy throws a stone with u=20m/s at some angle θ to vertical so that it reaches him. Find θ and also time of flight...
:)
-
UP 0 DOWN 0 0 19
19 Answers
33Yes thats a good method... and it thought it was obvious..so i didn't give this method.. [3]
same method as told by nishant bhaiya can be helpful in below post's question:
http://targetiit.com/iit_jee_forum/posts/no_topic_1407.htm
62hmm... u guys din observe one simple thing!!!!! I am a bit surprised! :O
The net acceleration is √102+42 = √116 = 2√29
initial velociy is 20
s=0
ut=1/2at2
t=2u/a = 20/√29
In normal case, We throw the ball vertically upwards (in the direction of force)
Here what is the direction of external force??? (that wud have been the answer .. simple!!)
I thought this is why you had given this one priyam...
I dont know which one is more enlightening!
33the method was based on superposition.. very helpful in physics as U all know.. ;)
33
all those vectors are displacement vectors due to all the causes..
velocity.. horizontal acc, vertical acc..
so θ is clear from that rt triangle.. tanθ=2t2/5t2.. tanθ=2/5
also by pythagoras theorem
(2t2)2+(5t2)2=(20t)2
as t≠0
therefore..
4t2+25t2=(20)2
t=20/√29
:)
1horizontal disp. 0=vsinθt-1/2at2 where θ is with the vertical
t=0 or 2vsinθ/a
also time of flight = 2vcosθ/g
equatin we get tan θ=2/5
put sinθ=2/√29 to get time of flight
...same old layman method but i wud love to see what method u have to show
33Again cheers subhas..
can u give an outline to ur son.. plzzz
62hmm....
is it mockery of projectiles or those who cant solve it in 1 minute?
33@ PHILIP yes velocity should have been given edited there u is 20m/s
have 2 find time period and θ
I wanted to give the method but 100% confident answer first..
:)
1let time of flight be t
u sinθ = 4t
u cosθ = 10(t/2)
solving
θ = tan-1(4/5)
:-) * * * * * (-:
1oops
yes init. vel.l must be given...
but θ is sky correct i think so
i had measured it wid the horizontal
but how 4/5 ??
1well, θ is asked with the vertical,
so θ = tan-1(4/5)
& for time of flight, initial velocity must be given,
i think so,,,,,,,,,,
1nahi yaar! theta wala tukka thiik hai mera .. priyam confirmed..
but u is surely wrong :P
33Nooo....
it's just an example :P
(aur koi example yaad hi nahi aya :P)
it is to show one method... :)