Question

body is projected with velocity u making an angle with the horizontalios velocity when it is
perpendicular to the initial velocity vector is
1) u sin
2) u cote
3) u tan
4) u cos e
​

Answer 1

Answer:

U cote second is the right answer

Answer 2

Answer:

shown in the figure, when the velocity vector v at point P becomes perpendicular to initial velocity vector u,

the resolved. vertical component of v makes angle x with the velocity vector v.

hence we have, begin mathsize 12px style v subscript vertical line vertical line end subscript over v subscript perpendicular space equals space fraction numerator u space cos space x over denominator left parenthesis u space sin space x right parenthesis space t space minus begin display style 1 half end style g t squared end fraction space equals space tan x end style.................(1)

Eqn.(1) be re written as

begin mathsize 12px style u space cos squared x space equals space open parentheses u space sin squared x close parentheses space t space minus open parentheses 1 half g space sin x close parentheses space t squared

t squared space minus space open parentheses fraction numerator 2 u over denominator g end fraction sin space x close parentheses t space plus fraction numerator 2 u over denominator g end fraction cos space x space c o t space x space equals space 0 space.......... left parenthesis 2 right parenthesis end style

eqn.(2) has solution :- begin mathsize 12px style t space equals space u over g open curly brackets space sin space x space plus-or-minus space square root of 1 space minus space c o t squared x end root close curly brackets end style

we omit the solution with -ve sign, because physically it is not possible

( velocity vector v becomes perpendicular to initial velocity vector u, after the instant t = (u sin x)/g , when it reaches the maximum height, not before)

Hence at that instant, begin mathsize 12px style t space equals space u over g open curly brackets space sin space x space plus space square root of 1 space minus space c o t squared x end root close curly brackets end style velocity vector becomes perpendicular to the initial projection velocity vector.

second is the correct answer

Please follow me...

...