A waterskier skis over the ramp (with a base of 5 m and a height of 1 m) at a speed of 12 m/s. How fast is she rising as she leaves the ramp?

Question

I am having trouble with this question as I do not understand how I would apply derivatives to this question. What is the process/steps for solving this word problem? Thank you very much!

Answer 1

Mary,

 

This is a trivial Algebra problem that requires no Calculus because the only question involves the one moment in time when the skier leaves the ramp.  Then calculus is overkill to answer this one question.  However, if they went on to ask questions about the skiers subsequent trajectory (max height reached?, time of impact on ground?, distance traveled horizontally at impact?), Calculus would be needed.  I'll do it both ways so you won't stay confused.

 

Let      x, y  be a coordinate system (x = horizontal distance, y = vertical distance, both measured in meters).

    Ski ramp be a RIGHT triangle with vertex at the origin, and (x,y) = (5,0) and (5,1) be the other 2 vertices.

    Triangle's hypotenuse = √[(5)² + (1)²] = √[26] = 5.1 meters

    Θ be the acute angle of the ramp at the origin, tan Θ =y/x = 1/5 = 0.20, so Θ = 11.54^o, but we don't need it.

    SIN Θ  = y/hypotenuse   = 1/5.1 = 0.196. This is what we need, not the angle itself.
    COS Θ = x/hypothenuse = 5/5.1 = 0.980. we'll use this in the Calculus solution.

    |V_o| = magnitude of the velocity or speed at end of the ramp, along the hypotenuse.

    t = time of flight after leaving the ramp.

    g = downward deceleration due to gravity at the earth's surface (= 9.8 m/sec²)  

 

Consider the waterskier a particle that stays in contact with the ramp until the ramp ends and then continues in free flight. So x and Y distances will be functions of t in free flight (t=0) when ramp ends.

 

(a) Algebraic solution of the only question you posed:

 

Just decompose the speed along the inclined ramp's hypotenuse along its x & y coordinates.  Actually, your question only cares about the vertical, y direction.

 

So V_y(t=0) = |V_o| SIN Θ = (12 m/s) (0.196)  =   2.352 m/s  = the rising speed

     V_x(t=0) = |V_o| COS Θ = (12 m/s) (0.980) = 11.760 m/s  = the horizontal speed

 

NOTE: These equations and velocity components are only correct at time t=0 while the downward vertical force of gravity is balanced or offset by the ramp.  During free flight, gravity will keep decreasing V_y until it hits zero at impact.  Gravity does not affect V_x.  Also note, that if you square V_y and V_x separately, add them, and then take the square root you will get back the 12 m/s speed.  So the triangle works for both distance and speed at the end of the ramp.

 

(b) Using Calculus:

 

From Physics, the free-flight equations for particle motion AFTER leaving the ramp are:

   x(t) =  x(0) + (V_x) t  =  X(0) + [|Vo| COS Θ] t  = 5 + 11.760 t

   y(t) =  y(0) + (V_y) t - 0.5 g t² = y(0) + [|Vo| SIN Θ] t - 0.5gt² = 1 +  2.352 t - 4.9 t².  g= gravity = 9.8 m/s^2.

 

The vertical velocity at any time t is dy(t)/dt = 2.352 - 9.8 t.  At t=0 it becomes 2.352 m/s as in part (a)

And dx(t)/dt = Vx(t) = 11.760 doesn't change because it's unaffected by the vertical-only gravity force.

 

To answer other questions about the particle trajectory:

Set dy(t)/dt = 2.352 - 9.8 t = 0 for the time at which max height occurs, t = 2.352/9.8 = 0.24 seconds

Use that time in y(t=0.24) formula to get the maximum height reached.

Set y(t*) to zero, solve the quadratic polynomial for t*, the time of impact.

Set t=t* in x(t) to find the horizontal impact point.

 

NOTE.  So you see that Calculus is a lot more flexible for answering many questions, but overkill for the one you asked.

 

I hope this helps.  Read it carefully.