Two balls are projected at different angles from the same place and with the same initial speed of 50 m/s. Both balls have the same range of 216 m. The difference in their times of flight is close to
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Oh, I don't knew this formula! Thanks!
How did you value of angle from
sin 2 θ = 2 5 0 2 1 6 ?
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2 5 0 2 1 6 ≈ 2 3 = s i n 6 0 ° = s i n 1 2 0 °
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Oh, so you approximated it! How can I know which angle to approximate? It is very difficult for me.
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@Vinayak Srivastava – Just practice more and more and it will become normal. :)
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@Aryan Sanghi – Ok, thanks! Btw, I think this is in 10th or 11th?
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@Vinayak Srivastava – Class doesn't matter, I knew this equation in 8 t h (try my profile for similar problems). I'll recommend you to try proving the equation yourself. If you're unable to, mention me, I'll send you a link. :)
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@Aryan Sanghi – @Aryan Sanghi , did you give NSEJS? I am planning to give it, but I haven;t started any preparation.
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@Vinayak Srivastava – Yes, I gave. NSEJS really requires speed and knowledge. You should really give it, it will give you a really strong motivation to learn more and more and..... Rajdeep Singh Dhingra bhaiya, he is of my brother's time who reached IJSO, so I think he'll explain you better. He solved and posted questions of extreme difficulties which I could solve only now, at your age. He is really an inspiration.
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@Aryan Sanghi – Can you tell important chapters as very less time is left? I really want to qualify and I will prepare from today if I know what to do. I don't like reading the NCERT.
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@Vinayak Srivastava – You should not read NCERT(of class 9th and 10th Science). Try book Pearson-For IIT-JEE to get a start to the chapters. NSEJS is a very basic version of IIT-JEE, so it will help. Complete it's 8 t h , 9 t h and 1 0 t h class versions with perfection, even if you're in 9 t h . Next, solve Previous year papers . When you're done, try my questions (or) Rajdeep Singh Dhingra's ones of previous time(2017 for me and 2015 for him). Complete these and then you could contact me if time is left.
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Find Range
Range = g u 2 s i n 2 θ
2 1 6 = 1 0 ( 5 0 ) 2 s i n 2 θ
s i n 2 θ = 2 5 0 2 1 6 ≈ 2 3
2 θ = 6 0 ° or 2 θ = 1 2 0 °
θ = 3 0 ° or θ = 6 0 °
Find time difference
time period(T) = g 2 u s i n θ
T 1 − T 2 = 1 0 2 ( 5 0 ) s i n ( 6 0 ° ) − 1 0 2 ( 5 0 ) s i n ( 3 0 ° )
T 1 − T 2 ≈ 3 . 6 s e c