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A particle is dropped from the top of a tower h meter hingh and at the same moment another particle is projected upwards from the bottom they meet when the upper one has descended a distance h/n. show that the velocities of the two when they meet are in the ratio 2:(n-2) and that the initial velocitiy of the particle projeceted up is [(1/2)ngh]1/2

shubham , 6 Years ago
Grade 12
anser 1 Answers
Ambareeshan R

Last Activity: 6 Years ago

for 1st particle
h/n= 1/2(gt2)------>eq.1
 
for 2nd particle
(h – h/n)=ut – 1/2(gt2)------->eq.2
 
Add 1 and 2:
h=ut or t=h/u----->substitute in 1.
 
h/n=1/2g(h2/u2)
u2=gnh/2
u=(gnh/2)1/2
 
now for 1st particle
V12=2g(h/n)
V22= u2 – 2g(h – h/n)
V22 = 1/2(ngh) – 2gh(h-h/n)
V22 =1/2(ngh) – 2gh(1-1/n)
 
V22=gh(n/2 – 2(1-1/n))
 
V22 =gh(n/2 – 2 + 2/n)=  gh((n – 2)2/2n)
 
TAKING RATIO
 
V12 /V22 = (2gh/n) / (gh(n – 2)2 /2n)
 
V12/V22 = 4gh/gh(n – 2)2
V12/V22 =2/n – 2
 
 
 
 
           
 
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