  Question

# Can you tell the solution of the question below ??? " A ball is projected with a speed of 20 m/s at an angle of 3o degree from a point on the top if a very high tower . the time after which it velocity becomes perpendicular to the velocity of projection ( take g = 10 m/s^2 ) "

Solution

## Let the horizontal component be x and vertical component be y. X = v(cos theta)t Y= v(sin theta)t-1/2 (gt^2) Initially the slope of the ball to the horizontal =sin theta/cos theta =tan theta At any time “t” the slope of the ball =v (cos theta)t/v (sin theta)t-1/2 (gt^2) =dy/dx If the direction of the ball is perpendicular to the initial direction then: (dy/dx)*(tan theta) = -1 So, V(sin^2theta)t-1/2 (gt^2)/V (cos theta )t*(tan theta ) = -1 Hence :- V (sin^2theta )-(gt*sin theta ) = -(Vcos^2theta ) Thus:- V (sin^2theta +cos^2theta) =gt (sin theta ) V=gt (sin theta ) So , coming back to the question Time at which the velocity is perpendicular to the initial velocity is given by T=v/g (sin theta ) Replacing them with numerical values we get:- T=20/10*sin 30 T=20/10*0.5 T=20/5=4 Therefore after 4 seconds the initialvelocity would be perpendicular to the velocity.  Suggest corrections   