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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
+1 vote

The velocity of projection of an oblique projectile is $(6i +8j)m/s$ . The horizontal range of projectile is $( g=10 m/s)^2$

$\begin{array}{1 1}(A)\;4.9 \;m\\(B)\;9.6\;m \\(C)\;19.6\;m\\(D)\;14 \;m \end{array} $

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1 Answer

+1 vote
Answer :$(B)\;9.6\;m$
$\overrightarrow{v} = 6 \hat {i} +8 \hat {j}$
Comparing with
$ \overrightarrow {v} = v_x \hat i = v_y \hat j$ we get,
$v_x = 6 ms^{-1}$
$v_y= 8 ms^{-1}$
Also, $v^2 =v_x^2 +v_y^2$
$\qquad= 36+64 =100$
or $v= 10 ms^{-1}$
$\sin \theta = \large\frac{8}{10}$ and $ \cos \theta= \large\frac{6}{10}$
$R= \large\frac{v^2 \sin 2 \theta}{g} =\frac{2v^2 \sin \theta \cos \theta}{g}$
$R= 2 \times 10 \times 10 \times \large\frac{8}{10} \times \frac{6}{19}\times \frac{1}{10} $$\;m$
$\qquad=9.6 \;m$
answered Jul 14, 2014 by meena.p
edited Aug 13, 2014 by meena.p

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