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A particle is projected at an angle 30 degrees. At a partciular height 'h'. velocity of particle is (9i + 3j)m/s. Find 'h'
IT IS VERY EASY. FIRST FIND OUT THE SPEED AT WHICH THE PARTICLE IS THROWN .
SPEED= (9^2 + 3^2)^(1/2) = 3*10^(1/2)
THE ANGLE IS ARCTAN(9/3)= ARCTAN 3. NOW DROP A PERP FROM THE HEIGHT REACHED AND SOLVE THE PROBLEM USING A LITTLE TRIGONOMETRY.
DO TELL ME IF U NEED MORE ASSISTANCE.
Hi Abishek,
The horizontal component of the velocity remains unchanged.
So vcos30 = 9.
Hence v = 18/√3. = 6√3.
Now in the vertical direction use eqn (final vel)2 - (intial vel)2 = 2ah.
So 3^2 - (6√3sin30)^2 = 2*(-g)*h
which would give 18 = 2gh.
Or h =9/g.
Where g is the acceleration due to gravity.
All the best.
Regards,
Ashwin (IIT Madras).
Thanks a lot Ashwin!
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