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If the initial velocity of a projectile is doubled while the angle of projection is unchanged, the horizontal range becomes:
Atwice the original range
Bfour times the original range
Chalf the original range
Dthe same as the original range
Answer & Solution
Correct answer: B. four times the original range
1. The horizontal range is $R=\dfrac{v_0^2\sin 2\theta}{g}$.
2. The angle $\theta$ is held constant, so $\sin 2\theta$ and $g$ do not change.
3. That leaves $R$ proportional to $v_0^2$.
4. Doubling $v_0$ multiplies $v_0^2$ by four.
5. So the range becomes four times the original. Answering 'twice' treats the relation as linear in $v_0$.
_Source: NECTA ACSEE 2023 Physics 131/1, Question 2: Mechanics (Projectile Motion)_