To find minimum velocity.

In X direction, acceleration a = 0

Distance traveled, s_x = u_xt + 0 = u_xt = 3 (given)

                          u_x = 3/t.............................................................. equation 1

 

In Y direction, acceleration = -10 m/s²

Distance traveled, s_y = u_yt + (at²)/2 = u_yt - (10t²)/2 = u_yt -5t² = 4 (given)

                          u_y = (4 + 5t²)/t....................................................equation 2

 

Now velocity u = √(u_x² + u_y²) = √[(3/t)² + ((4+5t²)/t)²] = √[(25/t²) + (25t²) + 40]...equation 3

 

For min vel, differentiate equation 3 with respect to time and equate to zero.

this gives t = 1.

 

Putting t = 1 in equation (1) and (2), 

 u_x = 3

 u_y = 9

 

u = √(u_x² + u_y²) = √(3² + 9²) = 3√10.

 

Here's your answer.