A juggler tosses a ball in the air with the initial speed u. At the instant reaches its maximum height he tossed up the second ball with the same initial speed teh two balls will collied at height

Asked by Eswer Saran | 22nd Oct, 2013, 06:42: PM

Expert Answer:

When the first ball is thrown up, we have
 
u = u, v = 0, a = -g, s = H (maximum height)
 
Thus, we have
 
v2 = u2 - 2gH
0 = u2 - 2gH
H = u2/2g                                            ...... (1)
 
Now, when this ball comes down from maximum height. At the same time second ball is thrown up. Let the two balls collide at height h from the ground after time t.
 
Thus, for first ball
 
H - h = 0 + ½gt2                                   ...... (2)
 
And, for second ball
 
h = ut - ½gt2                                        ...... (3)
 
From (2) and (3), we get
 
H = ut                                                  ...... (4)
 
Thus, from equation (1) and (4), we get
 
t = u/2g                                               ...... (5)
 
Substitute equation (5) in equation (3)
 

Answered by Romal Bhansali | 11th Nov, 2013, 10:10: AM

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