2 identical balls are thrown upwards in the same vertical line at an interval of 2seconds with a speed of 40m/s each. Find the height at which they collide.

Asked by tanishq gupta | 29th Jul, 2012, 09:41: AM

Expert Answer:

Clearly it can be known from the situation that they will collide when first ball will be coming down and the second ball is still moving up.
Time taken by first ball to reach maximum height is T:
(using first equation of motion)
0 = 40 +(-10)T                  (here acceleration due to gravity is taken as negative as downward direction is taken as negative and  
                                           upward direction as positive)
T= 4 s
Height reached by first ball:
H = (40)(4) - 1/2(10)(4)2                     (using 2nd equation of motion)
H = 80 m
height reached by second ball in 2 sec (because the second ball is thrown after 2 sec of first ball)
h = 40(2) - 1/2(10)(2)2                             (using second equation of motion)
h = 60 m
Speed of second ball after 2 sec, v:
v = 40 - (10)(2) = 20 m/s
Thus now the two balls will colide somewhere between H and h
Let them collide at x height from h.
and let time taken from that time be t
Thus for first ball:
(H-x) = 1/2(10)(t)2
and second ball:
h+x = 20(t) - 1/2(10)(t)2
60+x = 20t -5(t)2
now x can be found by solving this quadratic equation
and the total height will be:  60 + x

Answered by  | 29th Jul, 2012, 03:37: PM

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