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  •  A ball is dropped from a high-rise platform at   starting from rest. After

 A ball is dropped from a high-rise platform at \mathrm{t}=0  starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed  v. The two balls meet at  t=18 \mathrm{~s}. What is the value of v ? (Take g=10 \mathrm{~m} / \mathrm{s}^2 )

Option: 1

75\ \frac{m}{s}


Option: 2

80\ \frac{m}{s}


Option: 3

120\ \frac{m}{s}


Option: 4

None of the above.


Answers (1)

For ball 1:

  h=\frac{1}{2}gt^2

 = \frac{1}{2}\times10\times{18}^2=1620\ m

For ball 2: 

h=1620m,\ u=v,t=18-6=12s

Therefore,  h=ut+\frac{1}{2}gt^2

Or, 1620=12v+\frac{1}{2}\times10\times{12}^2

Or, 12v=900

Or, v=75\ \frac{m}{s}

Posted by

Ramraj Saini

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