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The range of projectile which is launch at an angle of $15^{\circ}$ with the horizontal is $1.5 \;km$ what is the range of the projectile if it is projected at an angle of $45^{\circ}$ to the horizontal ?

$\begin{array}{1 1}(A)\;1.5 km\\(B)\;3.0\;km \\(C)\;6.0 km\\(D)\;0.75 \;km \end{array} $

1 Answer

$\large\frac{R_2}{R_1} = \frac{\sin 2 \theta_2}{\sin 2 \theta_1} =\large\frac{\sin 90^{\circ}}{\sin 30^{\circ}}$
$\qquad= 2$ (near value is 1.5)
Hence A is the correct answer.
answered Jul 14, 2014 by meena.p

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