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Derive an expression for maximum height ...

Derive an expression for maximum height of a projectile .

Text Solution

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Using `v^2=(v_0sintheta)^2-2gh`
for vertical motion, and setting
v=0 for h=H
We get 0`=v_0^2sin2theta-2gH`
`:.` Maximum height `H=(v_0^2sin^2theta)/(2g)`
Hence `Hpropv_0^2` for a given angle of projection.
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