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For a projectile launched along a smooth inclined plane, the range along the incline is: $x = \dfrac{2u^2\sin(\theta-\alpha)\cos\theta}{g\cos^2\alpha}$, where $\alpha$ is the incline angle and $\theta$ the launch angle from horizontal.
When the projectile is fired along the incline: $\theta = \alpha$, so $\sin(\theta-\alpha) = 0$... Using the general formula and applying both cases gives $x_1 : x_2 = \mathbf{1:\sqrt{3}}$.
Given that $0 < x < y < 1$, compare the two quantities:
Quantity A: $1 - y$
Quantity B: $y - x$
We know $0 < x < y < 1$. Let's test specific values to see if the relationship is fixed.
Example 1: $x = 0.1,\ y = 0.2$.
Quantity A: $1 - 0.2 = 0.8$. Quantity B: $0.2 - 0.1 = 0.1$. A > B.
Example 2: $x = 0.1,\ y = 0.9$.
Quantity A: $1 - 0.9 = 0.1$. Quantity B: $0.9 - 0.1 = 0.8$. B > A.
Since the relationship changes depending on the values of $x$ and $y$, it cannot be determined from the information given.
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