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Grade: 11
        
From the top of a tower,a stone is thrown up and it reaches the ground in time t1.A second stone is thrown down and it reaches the ground in t2 seconds .A third stone is released from rest and it reaches the ground in time t3. Give the relation btw t1,t2,t3?
5 years ago

Answers : (1)

Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							
Hello
Well i’m assuming the stone was throne down with same speed at in the previous case it was thrown up.
0 = h – (g t1^2)/2 + t1v
0 = h – (g t2^2)/2 – t2v
0 = h – (g t3^2)/2
=>substituting h
0 = 1/2 (g (-t1^2 + t3^2) + 2 t1 v)
0 = ½ (g (-t2^2 + t3^2) – 2 t2 v)
=>eliminating v
g (t1^2 - t3^2)/(2 t1) = g (t3^2 - t2^2)/(2 t2)
t3 = √(t1*t2)
so t3 is GM of the other two
Thanks & Regards
Arun Kumar
Btech IIT delhi
Askiitians Faculty
5 years ago
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