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!