Dimensional Analysis Type Promblem

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DoktorD
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Dimensional Analysis And Projectile Motion Problem

1.
Textbook problem 36 with Figure P4.56: a projectile fired up an incline at angle phi with speed v0 at angle theta0 to the horizontal, showing the range d up the slope and the path of the projectile.




The work I have shown below is my attempts at doing the steps shown to get that equation in the problem, which is what the "answer" is. I need to find the work. Am I on target?

And I am totally clueless as to how I get the second answer. I know that [tex]\theta[/tex] would equal 45 degrees right? Optimal launch angle disregarding air resistance. But I'm not sure.



3. The Attempt at a Solution
Handwritten derivation of projectile range d = 2 v0^2 cos θ0 sin(θ0 − φ) / (g cos^2 φ), with x and y motion equations, right-triangle diagrams of θ and φ, and substitution of t into the y equation.
 
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Work:Given: v_0 = 25 m/s, g = 9.81 m/s^2, h = 10 m We want to find the optimal launch angle \theta 1. Convert the initial velocity to SI units: v_0 = 25 m/s = 25 * (1000 m/1 km) / (3600 s/1 hr) = 6.94 km/hr 2. Convert the gravity to SI units: g = 9.81 m/s^2 = 9.81 * (1000 m/1 km) / (3600 s/1 hr)^2 = 0.27 km/hr^2 3. Convert the height to SI units: h = 10 m = 10 * (1000 m/1 km) = 10 km 4. Calculate the optimal launch angle: \theta = tan^{-1} (\frac{2v_0^2}{gh}) = tan^{-1} (\frac{2*(6.94 km/hr)^2}{0.27 km/hr^2 * 10 km}) = 45.024 degrees Answer:Optimal launch angle \theta = 45.024 degrees