Question

A particle is projected from ground at an angle 45 with the initial velocity 20√2 m/s .The magnitude of average velocity in a time interval from t=0 to t=3 is in m/s is ​

Answer 1

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Answer 2

Hello mate...

average velocity = (total displacement)/(time taken)

max height = u2cos245/2*g = 20 m

Half range = u2sin2*45/2*g = 40 m

time taken = 2usin45/g = 4 sec

total displacement = \sqrt{20^{2} + 40^{2}} = 20\sqrt{5}

so average velovity = 20\sqrt{5}/4 = 5\sqrt{5} m/s

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