Question

A body is thrown with a speed of 20m/s at an angle of 60 with the horizontal find the time gap between the two positions​

Answer 1

Question: A body is thrown with a speed of 20m/s at an angle of 60 with the horizontal find the time gap between the two positions​.

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 60

T=20/10 × √3/2

T=2 × 0.866 =1.732

Therefore after 1.732 seconds the initialvelocity would be perpendicular to the velocity.