2 Dimensional Motion

Classical Mechanics Level 3

If R R is the horizontal range for an inclination and h h is the maximum height reached by the projectile, then the maximum range is given by __________ \text{\_\_\_\_\_\_\_\_\_\_} .

More questions??

R 2 8 h − 2 g h \frac { { R }^{ 2 } }{ 8h } -2gh R 2 8 h + 2 h \frac { { R }^{ 2 } }{ 8h } +2h None of these R 2 8 h − 2 h \frac { { R }^{ 2 } }{ 8h } -2h

Anandhu Raj by Anandhu Raj , 23, India

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1 solution

Anandhu Raj
Feb 11, 2016

We know range, R = u 2 sin ⁡ 2 θ g = u 2 2 sin ⁡ θ cos ⁡ 0 g R=\frac { { u }^{ 2 }\sin { 2\theta } }{ g } =\frac { { u }^{ 2 }2\sin { \theta } \cos { 0 } }{ g }

⇒ R 2 = 4 u 2 sin ⁡ 2 θ cos ⁡ 2 θ g 2 = u 2 sin ⁡ 2 θ 2 g × 2 × 4 u 2 cos ⁡ 2 θ g \Rightarrow { R }^{ 2 }=\frac { 4{ u }^{ 2 }\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } }{ { g }^{ 2 } } =\frac { { u }^{ 2 }\sin ^{ 2 }{ \theta } }{ 2g } \times 2\times \frac { 4{ u }^{ 2 }\cos ^{ 2 }{ \theta } }{ g }

⇒ R 2 = h × 8 u 2 cos ⁡ 2 θ g ⇒ R 2 8 h = u 2 cos ⁡ 2 θ g \Rightarrow { R }^{ 2 }=h\times \frac { 8{ u }^{ 2 }\cos ^{ 2 }{ \theta } }{ g } \quad \Rightarrow \frac { { R }^{ 2 } }{ 8h } =\frac { { u }^{ 2 }\cos ^{ 2 }{ \theta } }{ g }

⇒ R 2 8 h + 2 h = u 2 cos ⁡ 2 θ g + u 2 sin ⁡ 2 θ g = u 2 g ( cos ⁡ 2 θ + sin ⁡ 2 θ ) \Rightarrow \frac { { R }^{ 2 } }{ 8h } +2h=\frac { { u }^{ 2 }\cos ^{ 2 }{ \theta } }{ g } +\frac { { u }^{ 2 }\sin ^{ 2 }{ \theta } }{ g } =\frac { { u }^{ 2 } }{ g } \left( \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right)

⇒ R 2 8 h + 2 h = R m a x \Rightarrow \boxed{\frac { { R }^{ 2 } }{ 8h } +2h={ R }_{ max }}

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