IIT JEE , AIEEE Forum - Physics , Mechanics

rathin_123

find the average velocity of a particle between the instants it crosses half the maximum height.it is projected with a speed u at an angle thita with the horizontal.

SCIENTIST135

what do you mean by half half ?

madman

ucos@

SCIENTIST135

(VELOCITY)_AV =( V₂ - V1) /₂ NOW V₁ = u

VELOCITY AT ANY HEIGHT,
LET P BE THE POINT ON THE PROJECTILE'S TRAJECTORY AND AT A HEIGHT h.
and let V₂ be the velocity at that instance. let @ be the angle that V₂ makes with the horizontal velocity.

now, V₂COS@ = V₁COS$ (1) let THITA = $ (BECUASE, HORIZONTAL VELOCITY COMPONENTS ARE EQUAL)
(V₂SIN@)² = (V₁SIN$)² - 2gh. (2) (FROM V₂ = V₀² -2GH)

SQ EQ 1 AND ADDING 1,2 WE GET,

V₂ = (V₁ - 2gh)^1/2 HENCE IF MAX HEIGHT IS 'H' THEN IN THIS CASE,h=H/2

HENCE, V₂ = (V₁ - gH)^1/2

now average = (V2 - V1)/2 = [(u - gH)^1/2- u] / 2.

RATE ME WITH HAT IF I AM RIGHT OR CLEAR !!!!

madman

he has asked the average velocity between the times when it is at a height H/2
not between 0 and H/2

SCIENTIST135

NO ITS INITIAL VELOCITY IS u AND AVERAGE VELOCITY ITSELF MEANS VELOCITY FROM 0 TO H/2. i.e HALF OF FINAL - INITIAL.

SCIENTIST135

OKAY I AM SORRY, MISCONCEPTION !!!!! YOU ARE RIGHT MADMAN.

SCIENTIST135

EDITED : AS VERTICAL VELOCITY AVERAGE IS ZERO AND HORIZONTAL VELOCITY IS CONSTANT i.e ucos(thita). HENCE AVERAGE VELOCITY IS ucos(thita) IN HORIZONTAL DIRECTION.

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