# Use of direction of downward velocity in equation of motion

[NOTE: I am not asking anyone to solve the question below but to point out where I might be wrong]

Q- A skier jumps from a horizontal track and lands on a steeper track with a launch angle of ∅=11.3°(anticlockwise from $$x$$-axis) and a velocity $$u$$=10m/s. The steeper track has a slope or ∆=9°(angle clockwise from $$x$$-axis). How far below the launch level, does the skier land?

My approach

1. I decide to get the required time to fall down from the same level as the launch level on the other end to the steeper track.

2. I first frame up an equation of the steeper track like so:

$$y = -tan(\Delta)x$$

1. The skier's launching velocity can be resolved into components along the $$x$$ and the $$y$$ axes. The $$x$$-component remains constant and has a value of,

$$u_x = ucos(\phi)$$

1. Thus, it can be said that,

$$x = u_x t$$ (When the skier lands after time t)

1. Therefore,

$$y = -tan(\Delta)u_x t$$

1. Also, from the y-component of the velocity (say u_y),

$$y = -u_y t - 0.5gt^2$$

(Due to the case of a projectile motion, the $$y$$-velocity on the launch level when the skier will be falling will be the same but in the opposite direction i.e. downwards. Using such equations have always given me correct answers and also they seem reasonable to me(at least). For example, a negative displacement means that the object gets displaced downwards and I am considering downwards to be negative)

Equating equations from $$5$$ & $$6$$ and solving for $$t$$(neglecting $$t=0$$) yields:

$$t = \frac{2(tan(9°)u_x - u_y)}{g}$$

And $$t$$ comes out to be -ve, which isn't possible.

However, if the coefficient of $$u_y$$ in the equation from $$6$$ is +ve, the whole thing boils down to the required time(and the rest I can do). What am I understanding wrong?

Edit: A picture is always better than words. You are starting at the initial condition given by the green ball if you take $$u_{y_0}$$ negative, so you have to keep in mind $$x$$ is not $$a$$ as in the picture, and is not true $$\tan \Delta= \frac{h}{a} \neq \frac{y}{x}$$, but it is $$\tan \Delta= \frac{y}{x+b} = \frac{h}{a}$$. You can also see why starting with the blue initial condition ($$u_{y_0}$$ positive) solves your problems.
• Alright, what you said in your previous comment was misunderstood by me. The edit clarified it all. Thank you so very much for the explanation, however, just to clarify things out shouldn't it be $|tan(\Delta)| = \frac{y}{x-x_0}$(if I do use the falling velocity) ? – Sid Jun 5 at 14:23
• No problem! No, you have to use $\frac{y}{x+x_0}$ because your $x$ starts at the green point, and the horizontal side of your angle, $\Delta$, is $\Delta x$ in the picture. I will update the picture so you can see it. – Puco4 Jun 5 at 14:31