Question

A particle is pojected from the ground at an angle of theta with the horizontal with an initial speed of u. Time after which velocity vector of the projectile is perpendicular to the initial velocity is

Answer 1

Answer:

$initial \: vt \: velocity \: = usin \beta \\ initial \: hz \: velocity \: = ucos \beta \\ at \: tha \: point \: when \: initial \: velocity \\ is perpendicuar \: to \: velocity \: after \: \\ time \: t \\ hz \: velociy \: will \: alway \: remain \: same \: \\ ucos \beta = vsin \beta - - - eq1 \\ and \: in \: vt \: motion \: \\ vsin \beta = vsin \beta - gt \\ \frac{u}{tan \beta} \times sin \beta = usin \beta - gt \\ t = usin \beta (1 - \frac{1}{tan \beta} )$

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