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Q. A particle is dropped under gravity from rest from a height h (g= 9.8m/s2) and it travels a distance 9h/25 in the last second, the height h is
a)100m
b)122.5m
c)145m
d)167.5m

Sri Subhash Pathuri , 8 Years ago
Grade 11
anser 4 Answers
Vikas TU

Last Activity: 8 Years ago

As the particle is dropped then u = 0
        As we all know that g = 9.8 m/s^2.
Then from the third equation of motion=
S = ut+1/2 gt^2
S = ½ gt^2
where t is the total time taken
  • Distance covered in (t-1) sec = h -9h/25 = 16h/25
Solving (1.) and (2.)
T = 5s
Substituting the value of t in eq. 1
S = ½ *9.8 *25
S = 122.5 m
 
Approved
Ashwani Singh

Last Activity: 6 Years ago

Let h distance is covered in t second.
✓ h =1/2gt^2
Distance covered in t^th second=1/2g(2t-1)
✓9h/25=g/2(2t-1)
Solving both equations we get, h=122.5 m.
Navya khandelwal

Last Activity: 6 Years ago

h=1/2gt2.                         …......eq1
9h/25 = u+a/2(2t-1 .    …......eq2
9*1/2gt2=0+5(2t-1)
9t2-50t+25=0
Solve for t 
U will get t=5
Put t in eq 1 u will get h=125m
 
Kushagra Madhukar

Last Activity: 5 Years ago

Dear student,
Please find the solution to your problem.
 
As the particle is dropped then u = 0
As we know that g = 9.8 m/s2.
Then from the third equation of motion=
S = ut+1/2 gt2
S = ½ gt2
where t is the total time taken
  • Distance covered in (t – 1) sec = h – 9h/25 = 16h/25
Solving (1.) and (2.)
T = 5s
Substituting the value of t in eq. 1
S = ½ x 9.8 x 25
S = 122.5 m
 
Thanks and regards,
Kushagra
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