The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is _____ m.

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Solution

Q.
The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is _____ m. [2024]

Ans.
(16) $H_{\max} = 64 m$

$H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow \frac{H_{1 \max}}{H_{2 \max}} = \frac{u_{1}^{2}}{u_{2}^{2}}$

$\Rightarrow \frac{64}{H_{2 \max}} = \frac{u^{2}}{{(\frac{u}{2})}^{2}} \Rightarrow H_{2 \max} = 16 m$

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