To find minimum velocity.

Asked by Harshal Saptarshi | 5th Sep, 2010, 11:15: PM

Expert Answer:

In X direction, acceleration a = 0
Distance traveled, sx = uxt + 0 = uxt = 3 (given)
                          ux = 3/t.............................................................. equation 1
 
In Y direction, acceleration = -10 m/s2
Distance traveled, sy = uyt + (at2)/2 = uyt - (10t2)/2 = uyt -5t2 = 4 (given)
                          uy = (4 + 5t2)/t....................................................equation 2
 
Now velocity u = √(ux2 + uy2) = √[(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), 
 ux = 3
 uy = 9
 
u = √(ux2 + uy2) = √(32 + 92) = 3√10.
 
Here's your answer.

Answered by  | 6th Sep, 2010, 11:28: PM

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