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Question

The range of projectile when launched at an angle of $$15^{\circ} $$ with the horizontal is $$1.5\ km$$. What is the range of the projectile when launched at an angle $$45^{\circ}$$ to the horizontal 


A
1.5 km
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B
3.0 km
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C
6.3 km
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D
0.75 km
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Solution

The correct option is C $$3.0\ km$$

Here in first case, $$R_{1} = 1.5 , \theta = 15$$

$$R_{1} = \dfrac{u^{2}\sin (2\times 15) }{g}\Rightarrow  1.5 = \dfrac{u^{2}}{2g}$$

 $$ \Rightarrow \dfrac{u^{2}}{g}  = 3 $$

Now, for second case,

$$R_{2} = \dfrac{u^{2}\sin (2\times45) }{g}\Rightarrow   R_{2} = \dfrac{u^{2}}{2g} = 3 km$$


Physics

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