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A particle is projected vertically up and it reaches a 15 cm high mark at certain instant. After 4 seconds from this instant the particle crosses this mark again in its downward journey. What is the initial velocity of projection of this particle? $(g= 10 ms^{-2})$

$\begin{array}{1 1}(A)\;10 \sqrt 7 \\(B)\;6 \sqrt 2 \\(C)\;5 \sqrt 5 \\(D)\;6.7 \sqrt 2 \end{array} $

1 Answer

After 4 seconds , the particle returns to the same mark.
If its velocity when crossing 15 cm high mark is v then in 4s its displacement is zero.
Therefore $0= v \times 4 -\large\frac{1}{2}$$ \times 10 \times 4^2$
$v= 20 ms^{-1}$
If the initial velocity of projection is u, then
$v^2=u^2 +2as $ gives
$20^2 =u^2 -2(10)(5)$
$u^2= 700$
$u= 10 \sqrt 7 ms^{-1}$
Hence A is the correct answer.
answered Jul 25, 2014 by meena.p

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