What Happens When Krishna and SG Jump Together?

Let the time taken = t;

After time "t", velocities of

SG = (-50)i + (-gt)j

Krishna = (40)i + (-gt)j

{where

             i = unit vector in positive x-axis

             j = unit vector in positive y-axis

             g = acceleration due to gravity}

As their velocities are perpendicular at time "t";

dot product of their velocities must be equal to zero:

=> ( (-50)i + (-gt)j ) . ( (40)i + (-gt)j ) = 0

=> $g^{2} t^{2}= 2000$

=> $t^{2} =20$

I humbly request @shubhendra singh to correct the uploaded answer

I solved it as follows with a geometric approach,

If you draw vector diagrams of the velocities of two boys, y-component (let's call it 'v') will be same for both (gravity affects y-component of both of them) and x-component will be 50 and 40 in opposite direction.

Now, If the resultant of 'v' and 50 makes an angle theta with SG's x-velocity,

Then the resultant of 40 and 'v' should make the same angle theta with y-velocity 'v', for the two resultants to be perpendicular.

This gives, $\tan^{-1} \frac{v}{50} = \tan^{-1} \frac{40}{v}$

which gives v= $\sqrt{2000}$ and ultimately, since v=u+gt, and g=10,u=0, t= $\sqrt{20}$

Vector approach for solving this problem is the best way

We know that v = u + at so , In vector notation let i = +ve direction of x and j= -ve direction of y axis so speed of SG = v1 = 40i +10t speed of krishna = -50i+ 10t as the time when their velocities are mutually perpendicular, their velocities' dot product must be 0 by that we get, 100(t^2)-2000 = 0 It implies that t = sqrt(20) Hence our answer is 20