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The range of a projectile which is the launched at an angle of $15 °$ with the horizontal is $1.5 k m$ . What is the range of the projectile if it is projected of an angle $45 °$ to the horizontal ?

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$1.5 k m$

$3 k m$

$6 k m$

$0.75 k m$

answer is B.

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Detailed Solution

$R_{1} = \frac{u^{2} \sin (2 \times 15 °)}{g} = \frac{u^{2}}{2 g} R_{2} = \frac{u^{2} \sin (2 \times 45 °)}{g} = \frac{u^{2}}{g} ∴ R_{2} = 2 R_{1} = 2 \times 1.5 = 3 k m$

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