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A particle is projected with a velocity "v" note at an angle of 30 degree with the vertical.Find the average velocity for its time of flight.

IF POSSIBLE PLZ EXPLAIN WITH A FIGURE

 Since projectile motion is a two dimensional motion, we can find average velocity in two mutually perpendicular directions and then find the resultant average velocity. Projectile motion is a unique case of uniform acceleration and we can find components of average velocity by averaging initial and final values.

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When a particle is projected under gravity at a velocity u at an angle θ to the horizontal(neglecting air resistance)it follows the curve of a parabola.



The particle has an initial horizontal speed of ucosθ, which is unchanged throughout the motion.

Vertically the particle has an initial speed of usinθ. It falls under gravity and is accelerated downwards with an acceleration of g ms-1,where g = 9.8 ms-2 (approx.)

The time of flight is calculated from the vertical component of the velocity. It is the time it takes for the particle to go up, reach its maximum height and come down again. So this is twice the time to maximum height.

If the time to maximum height is t secs. Then the time of flight is 2t.

Consider motion up to maximum height. This is attained when the final velocity v = 0.

intial speed vertically upward is u sin θ

using., 

v= u +at

replace u by u sin θ

substituting from a= -g

where , 

v= 0 

0= u sin θ - gt

= t= u sin θ /g

= time of flight(2t) is 2u sin θ /g ..  

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