Distance of a Projective up a Ramp

[corbell777]

A projectile is fired with initial velocity v at angle a. It travels over a ramp of angle b. (a>b) Show that the distance the projectile travels up the ramp is d = 2v²cos(a)sin(a-b)/gcos²(b).

V_fymax = V_iymax + a_yt
0 = V_isin(a) – gt
t_ymax = V_isin(a)/g
t_xmax = 2V_isin(a)/g – V_isin(a-b)/g


Max x = 2V_i²cos(a)sin(a)/g – V_i²cos(a)sin(a-b)/g
Max y_{at xmax}= V_i²sin²(b)/2g

Then it should be d = sqrt(Max²x + Max²y).
But it’s not. What did I do wrong?

 

[galactus]

Try starting with the formula:

\(\displaystyle \displaystyle y=\tan(a)x-\frac{g}{2v^{2}\cos^{2}(a)}x^{2}\)

Now, set \(\displaystyle x=d\cos(b), \;\ y=d\sin(b)\)

\(\displaystyle \displaystyle d\sin(b)=d\tan(a)\cos(b)-\frac{g}{2v^{2}\cos^{2}(a)}\cdot d^{2}\cos^{2}(b)\)

Now, solve for d and it should simplify down to the needed formula.

Note it is a quadratic in d.

You can solve for d using the quadratic formula.