[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?


Notice the question says the skier jumps, which as far as I understand, it means it starts going up and then falls. This would correct your problem with the sign in equation from step 6.

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. enter image description here

  • $\begingroup$ As I explain in the next paragraph, the velocity for falling downwards is the same initial velocity during jump(Due to a full projectile and stuff). But as the skier is falling downwards, it is in the opposite direction i.e. downwards (which is -ve) $\endgroup$ – Sid Jun 4 at 21:43
  • $\begingroup$ But the equation is with the initial velocity. If you use the starting moment when it jumps you can use it positive. If you use it negative you have to bear in mind you are starting at a different x (when he falls he has already travelled forward) and time. $\endgroup$ – Puco4 Jun 4 at 21:59
  • $\begingroup$ Well I am applying the equations of motions only in the vertical directions and that is why I resolved the velocity into two components... $\endgroup$ – Sid Jun 5 at 8:59
  • $\begingroup$ 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) ? $\endgroup$ – Sid Jun 5 at 14:23
  • $\begingroup$ 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. $\endgroup$ – Puco4 Jun 5 at 14:31

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