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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.

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