Solving Frictionless Snow Physics Problem - How Far from Ramp?

[feedmeister]

Homework Statement

It's been a great day of new, frictionless snow. Julie starts at the top of the 60 ^\circ slope shown in the figure. View Figure At the bottom, a circular arc carries her through a 90 ^\circ turn, and she then launches off a 3.0-m-high ramp.
How far is her touchdown point from the base of the ramp?

Homework Equations

$$(1/2)mv_i^2 + mgh_i = (1/2)mv_f^2 + mgh_f$$

The Attempt at a Solution

I set mgh = 1/2 mv^2 + mgh
m(9.8)(25)=.5m(v^2) + m(9.8)(3)
I then got 4.08 m/s as the final velocity. But how do I figure out how far it goes if I don't know the acceleration or time?

Thanks,

 

[dynamicsolo]

Once the skier leaves the ramp, the rest is a ballistics problem. You found the speed at which they fly off; what is the angle to the horizontal at which they are launched? How do you figure out what happens to the skier after that?

 

[ogb p]

Julie, for presenting this interesting problem. I can provide a response by breaking down the problem and using relevant equations to solve it.

Firstly, let's define the variables in the problem. We have the initial velocity (v_i) of Julie at the top of the slope, the height (h_i) at which she starts, the final velocity (v_f) at the bottom of the slope, and the height (h_f) at which she lands. We also know that the slope is at a 60-degree angle and the ramp is 3.0 meters high.

Now, we can use the equation you mentioned, which is the conservation of energy equation, to solve for the final velocity (v_f). This equation states that the initial kinetic energy (1/2 mv_i^2) plus the initial potential energy (mgh_i) is equal to the final kinetic energy (1/2 mv_f^2) plus the final potential energy (mgh_f). In this case, we can assume that the initial potential energy is equal to zero since Julie starts at the top of the slope. Therefore, our equation becomes:

(1/2)mv_i^2 = (1/2)mv_f^2 + mgh_f

Substituting the values we know, we get:

(1/2)m(v_i^2 - v_f^2) = mgh_f

Solving for v_f, we get:

v_f = √(v_i^2 - 2gh_f)

Now, we can use this value of v_f to solve for the distance (d) travelled by Julie. We know that the distance travelled is equal to the average velocity (v_avg) multiplied by the time (t) taken to travel that distance. In this case, the average velocity is equal to (v_i + v_f)/2. Therefore, our equation becomes:

d = v_avg * t = (v_i + v_f)/2 * t

Substituting the values we know, we get:

d = [(v_i + √(v_i^2 - 2gh_f))/2] * t

Now, we need to find the value of t. We can use the equation of motion in the vertical direction, which states that the final velocity (v_f) is equal to the initial velocity (v_i) plus the acceleration (a) multiplied by the time (t). In