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A projectile is thrown at an inclination θ. A bird follows the path of the projectile maintaining a constant speed same as the initial speed of the projectile. Find the bird’s acceleration at its highest point?

Note by Ronak Pawar 3 years, 5 months ago

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Since velocity is constant we have acceleration always in the normal direction given by :

\(\frac { { v }^{ 2 } }{ R }\) where \(R\) is the radius of curvature.Now radius of curvature is given by

\(R=\frac { { v }^{ 2 }{ cos }^{ 2 }\theta }{ g } \) .Putting the values we have :

\(\boxed { a=\frac { { v }^{ 2 } }{ (\frac { { v }^{ 2 }{ cos }^{ 2 }\theta }{ g } ) } =g{ sec }^{ 2 }\theta }\)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

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`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestSince velocity is constant we have acceleration always in the normal direction given by :

\(\frac { { v }^{ 2 } }{ R }\) where \(R\) is the radius of curvature.Now radius of curvature is given by

\(R=\frac { { v }^{ 2 }{ cos }^{ 2 }\theta }{ g } \) .Putting the values we have :

\(\boxed { a=\frac { { v }^{ 2 } }{ (\frac { { v }^{ 2 }{ cos }^{ 2 }\theta }{ g } ) } =g{ sec }^{ 2 }\theta }\)

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