Projectile Motion

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A projectile is fired at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the maximum height reached? (g = 10 m/s²)

RankGuru · Accepted Answer

1. **Maximum Height Formula**: $H = \frac{u^2 \sin^2 	heta}{2g}$.
2. **Identify Variables**:
   - $u = 40$ m/s.
   - $	heta = 30^\circ$.
   - $g = 10$ m/s$^2$.
3. **Calculation**:
   - $\sin 30^\circ = 0.5$.
   - $H = \frac{(40)^2 	imes (0.5)^2}{2 	imes 10} = \frac{1600 	imes 0.25}{20}$.
4. **Final Value**: $H = \frac{400}{20} = 20$ m.
5. **Note**: The maximum height depends only on the vertical component of the initial velocity ($u \sin 	heta$).

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