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  • Help me understand! A ball is dropped from a high rise platform at 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=18s. What is the value of v? (take g=10m/s2)

A ball is dropped from a high rise platform at 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=18s. What is the value of v? (take g=10m/s2)

  • Option 1)

    75m/s

  • Option 2)

    55m/s

  • Option 3)

    40m/s

  • Option 4)

    60m/s

 

Answers (1)

best_answer

As we learnt in

2nd equation or Position- time equation -

s= ut +frac{1}{2}at^{2}

s ightarrow Displacement

u ightarrowInitial velocity

a ightarrowacceleration

t ightarrow time

 

-

 

 Distance moved in 18 sec by first half, S_{18} = \frac{1}{2}\times 10\times 18^{2} =90\times 18= 1620\ mt

Distance moved in 12 sec by second half, S_{12} = u\times 12+\frac{1}{2}\times10 \times12^{2}

S_{18} = S_{12}

1620 = 12v + 5 x 144

\Rightarrow v = 135-60 = 75 \ ms^{-1}


Option 1)

75m/s

Correct

Option 2)

55m/s

Incorrect

Option 3)

40m/s

Incorrect

Option 4)

60m/s

Incorrect

Posted by

Plabita

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