18. A projectile is fired at $60^{\circ}$ above the horizontal line with an initial velocity $v_{0}$. At which of the following angles the projectile will land at the same distance as it is landed in the first trial?
A. $20^{\circ}$
B. $30^{\circ}$
C. $40^{\circ}$
D. $45^{\circ}$

Question

Accepted Answer

∻Solution∻ ‖ a ⋮ The range of a projectile is given by the formula: $$R = \frac{v_0^2 \sin(2 heta)}{g}$$ where $R$ is the range, $v_0$ is the initial velocity, $ heta$ is the launch angle, and $g$ is the acceleration due to gravity. For the first trial, the angle is $60^{\circ}$, so we calculate the range as: $$R_1 = \frac{v_0^2 \sin(120^{\circ})}{g}$$. Since $\sin(120^{\circ}) = \sin(180^{\circ} - 60^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2}$$, we have: $$R_1 = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g}$$. ‖ ‖ b ⋮ To find another angle that gives the same range, we need to find an angle $ heta_2$ such that: $$\sin(2 heta_2) = \sin(120^{\circ})$$. The angles that satisfy this equation are: $$2 heta_2 = 120^{\circ} + n \cdot 360^{\circ} ext{ or } 2 heta_2 = 180^{\circ} - 120^{\circ} + n \cdot 360^{\circ}$$ for integer $n$. This gives us: $$ heta_2 = 60^{\circ} ext{ or } 30^{\circ}$$. Thus, one possible angle is $30^{\circ}$. ‖ ‖ c ⋮ The other angle that can give the same range is $90^{\circ} - 60^{\circ} = 30^{\circ}$. However, we also need to check the other options provided. The angle $20^{\circ}$ does not satisfy the equation, and neither do $40^{\circ}$ or $45^{\circ}$. Therefore, the only angle that will land at the same distance as the first trial is $30^{\circ}$. ‖ ∻Answer∻ ⚹ B ⚹ ∻Key Concept∻ ⚹ Projectile Motion: The range of a projectile is determined by its initial velocity and launch angle. The formula for range is $$R = \frac{v_0^2 \sin(2 heta)}{g}$$, where $g$ is the acceleration due to gravity. The angles that yield the same range are complementary angles. ⚹ ∻Explanation∻ ⚹ The projectile launched at $60^{\circ}$ has a complementary angle of $30^{\circ}$ that will yield the same range, as both angles satisfy the sine function relationship for projectile motion. ⚹

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