projectile motion

Question

Maximum height reached by the projectile of KN-003 is:

RankGuru · Accepted Answer

From KN-003, $u = 20 	ext{ m/s}$ and $	heta = 30^\circ$.
The formula for maximum height $H_{max}$ is:
$H_{max} = \frac{u^2 \sin^2 	heta}{2g}$
Substituting values:
- $\sin 30^\circ = 0.5 \implies \sin^2 30^\circ = 0.25$
- $u^2 = 400$
- $2g = 20$
$H_{max} = \frac{400 	imes 0.25}{20} = \frac{100}{20} = 5 	ext{ m}$.

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