Question

Show that the maximum height reached by a projectile launched at an angle 45degree is one quarter of the range

Answer 1

Solution :

⏭ Given:

✏ Angle of projection = 45°

⏭ To Prove:

✏ Max. Height = Range/4

⏭ Formula:

✏ Formula of Range and Max. Height in projectile motion is given by...

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

$\boxed{\sf{\purple{\large{H=\dfrac{u^2{\sin}^2\theta}{2g}}}}}$

⏭ Calculation:

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• Range

$\implies\sf\:R =\dfrac{u^2\sin2(45\degree)}{g}\\ \\ \implies\sf\:R=\dfrac{u^2\sin90\degree}{g}\\ \\ \implies\sf\:\red{R=\dfrac{u^2}{g}}$

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• Max. Height

$\implies\sf\:H=\dfrac{u^2{\sin}^2 45\degree}{2g}\\ \\ \implies\sf\:H=\dfrac{u^2(\dfrac{1}{\sqrt{2}})^2}{2g}\\ \\ \implies\sf\:\blue{H=\dfrac{u^2}{4g}}$

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• Relation between R and H

$\mapsto\sf\:\dfrac{R}{H}=4\\ \\ \mapsto\:\boxed{{\sf{\orange{\large{H=\dfrac{R}{4}}}}}}$

✒ Hence, proved !!