From point A at the foot of an inclined plane with angle θ =…
From point A at the foot of an inclined plane with angle $\theta = 30^\circ$, an object is thrown up the slope with speed $v_0 = 5\ \mathrm{m/s}$ at $\alpha = 45^\circ$ to the horizontal. (a) At what distance $l$ from A (along the slope) does it land on the inclined plane? (b) What is the greatest height it reaches above the horizontal?
Explanation of Method
A throw from the foot of an inclined plane.
Step 1: Where the path meets the slope
With A as origin, the path is $y = x\tan\alpha - \dfrac{g x^2}{2v_0^2\cos^2\alpha}$ and the slope is $y = x\tan\theta$. Setting them equal (for $x \ne 0$): $\tan\alpha - \tan\theta = \frac{g\,x}{2v_0^2\cos^2\alpha}$.
\[ x = \frac{2v_0^2\cos^2\alpha\,(\tan\alpha - \tan\theta)}{g}. \]
\[ x = \frac{2 \cdot 25 \cdot 0.5 \cdot (1 - 0.5774)}{9.81} \approx 1.077\ \mathrm{m}. \]
Step 2: (a) Distance along the slope
\[ l = \frac{x}{\cos\theta} = \frac{2v_0^2\cos\alpha\,\sin(\alpha - \theta)}{g\cos^2\theta}, \]
\[ l = \frac{1.077}{0.866}. \]
\[l \approx 1.24\ \mathrm{m}\]
Step 3: (b) Greatest height above the horizontal
The top of the path is at $x = \tfrac{v_0^2\sin\alpha\cos\alpha}{g} \approx 1.27\ \mathrm{m}$, beyond the landing point, so the object hits the slope while still rising. The highest point it actually reaches is the landing point (the free parabola's top, $\tfrac{v_0^2\sin^2\alpha}{2g} \approx 0.64\ \mathrm{m}$, is never reached):
\[ y_{\max} = l\sin\theta = 1.24 \cdot 0.5 \approx 0.62\ \mathrm{m}. \]
\[y_{\max} \approx 0.62\ \mathrm{m}\]
Result
\[\boxed{\,l \approx 1.24\ \mathrm{m}\,}\]
\[\boxed{\,y_{\max} \approx 0.62\ \mathrm{m}\,}\]
Remark
About the original solution. The formulas it used, $v_0^2\sin 2\alpha/(g\sin\theta)$ and $v_0^2\sin^2\alpha/(2g\sin\theta)$, are not valid for motion on an incline, so 5.1 m and 2.55 m are wrong. If the incline is ignored and the object simply lands on level ground, the range would be $v_0^2/g \approx 2.55\ \mathrm{m}$ and the height 0.64 m. If the 45° is measured from the slope instead of the horizontal, $l$ comes out the same, 1.24 m.