​Two balls are thrown vertically, one with an initial velocity twice that of the other .What will be the maximum  height of the ball thrown with greater velocity in terms of the other ?

Dear Student,

                   As we know that at maximum height velocity(v) = 0.
Let 'u' be the initial velocity of first ball,  so '2u' will be the initial velocity of another ball.
Now for first ball, using equation of motion,
                                                           v2 = u2 - 2gh                                        [ let 'h' be the height of first ball] u2 =  2gh            ............(1)     [ at max height, v=0].

          Now for 2nd ball,    
                                                                                    
                                         (2u)2 = 2gH                         [ given, initial velocity of 2nd ball=2u,H= max height of 2nd ball]                4u2  = 2gH        ...........(2)

                From equation (1) & (2)...
                                                                H  = 4h.

So, maximum height of 2nd ball is four times that of first ball.


Regards..

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