Question

A bullet fired at angle 30° with horizontal hits the
ground 3 m away. What can be the approximate
maximum range of this projectile?

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf R_{max}=3.6m }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Angle $\sf{(\theta)=30°}$

• Range (R) = 3m

$\large\underline\pink{\sf To\:Find: }$

• Maximum Range $\sf{(R_{max})=?}$

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Range (R) :

$\large{\boxed{\sf R=\frac{u^2sin2\theta}{g}}}$

$\large\implies{\sf u=\sqrt{\frac{Rg}{sin2\theta}}}$

$\large\implies{\sf u=\sqrt{\frac{3×10}{sin60°}} }$

$\large\implies{\sf u=\sqrt{\frac{30×2}{\sqrt{3}}} }$

$\large\implies{\sf u=\sqrt{\frac{60}{\sqrt{3}}}}$

$\large\implies{\sf u=\sqrt{35.2}}$

$\large\implies{\sf u ≈6m/s}$

For Maximum Range :-

$\large{\boxed{\sf R_{max}=\frac{u^2}{g}}}$

$\large\implies{\sf R_{max}=\frac{36}{10}}$

$\large\implies{\sf R_{max}=3.6m}$

$\large\red{\boxed{\sf R_{max}=3.6m }}$

Hence ,

Maximum range of this projectile is 3.6m

Answer 2

See the above answer to know the answer!!!