alleryThe speed of the falling package as it reaches the ground is found from:
A$V=\sqrt{V_x^2-V_y^2}$
B$V=\sqrt{V_x^2+V_y^2}$
C$V=\sqrt{V_x^2V_y^2}$
D$V=\sqrt{V_x^2/V_y^2}$
Answer & Solution
Correct answer: B. $V=\sqrt{V_x^2+V_y^2}$
1. The velocity has a horizontal component $V_x=v_0\cos\theta$, unchanged throughout the flight.
2. It also has a vertical component $V_y=v_0\sin\theta-gt$, which grows as the package falls.
3. These two components are perpendicular to one another.
4. The resultant of two perpendicular vectors is found by Pythagoras: $V=\sqrt{V_x^2+V_y^2}$.
5. Adding the components arithmetically ignores that they act at right angles and overstates the speed.
_Source: NECTA ACSEE 2023 Physics 131/1, Question 2: Mechanics (Projectile Motion)_
Keep practising Mechanics (Projectile Motion)
Browse all Mechanics (Projectile Motion) questions for ACSEE (Form 6), or explore more Physics topics.