What Should Be the Angle of the Slope for a Perfect Ski Jump Landing?

[lwc729]

Homework Statement

A ski-jumper comes off ramp with speed 10 m/s at angle 16° above horizontal. He hits the snow at point 30 m below point he took off. Slope is designed to make perfectly soft landing, what should be the angle of the slope where he hits? (neglect air resistance) (hint:(from book)find direction of velocity vector when he hits the snow)

Homework Equations

for ease of use we round gravity to -10m/s
Vx=(10m/s)*(cos 16°)
=9.6 m/s
Vy=(10 m/s)*(sin 16°)
=2.8 m/s

For x:
Vf=9.6 m/s
Vi=9.6 m/s
a=0

For y:
Δr=-30 m
Vi=2.8 m/s
a=-10 m/s

The Attempt at a Solution

That's as far as I can get without getting confused/overwhelmed. I'm thinking I need to find Δt so I can get the x displacement. But as far as what to do next?

 

[nure]

Find the y-component of the velocity when he hits the ground. Then from the x- and y-component of the velocity, find the angle the velocity makes with the horizontal.

As you have pointed out, the x- component of the velocity will be constant as long as you neglect drag. The y-component you can find by calculating the speed from the top of the trajectory and to 30 meters below the ski jump. Add those 30 meters to the vertical distance from the ski jump to the top of the trajectory to find the total distance in free fall. You then should be able to find the x-speed of the velocity when the man hits the snow.

 

[lwc729]

for the y component. i calculated the max height reached and added 30. came up with 30.392 m? to find the x component don't i need the Δt to multiply by the constant v to get distance traveled? But i can't figure out how to find time elapsed with given info. or am i going about this wrong?

 

[Bread18]

for the y component. i calculated the max height reached and added 30. came up with 30.392 m? to find the x component don't i need the Δt to multiply by the constant v to get distance traveled? But i can't figure out how to find time elapsed with given info. or am i going about this wrong?

What formulas are you using for this?

 

[nure]

You know that on the top of the trajectory the y-speed is zero. Then you use the formula

v_y=v_y,0+a_y*(t-t₀)

to find the time. And you already know the x-component for the velocity. Then find the y-speed at the bottom of the trajectory. You then have enough information to find the velocity when he hits the snow.

 

[lwc729]

i used the formula:
Δy=v_fy²-V_fi²/2a_y
=0²-(2.8 m/s)²/2(-10 m/s)
=.392 m
then i added the 30 m from takeoff to landing to get total. am i off in this thinking?

 

[lwc729]

"You know that on the top of the trajectory the y-speed is zero. Then you use the formula

vy=vy,0+ay*(t-t0)"

isn't there two unknowns in this equation? as I'm still unsure of the final velocity of y?

 

[nure]

You have to split the job into two parts(one for the way up and one for the way down). First find the vertical distance from the initial point to the top(as you have already done).

Then add this distance to the 30 meters to find the distance the man will be in free fall. Use one of the constant acceleration formulas to find the y-speed after (30+distance from ski jump to top of trajectory) in free fall. The initial speed will now be zero.

You then have tWo velocity components in the landing point. Then use simple trigonometry to find the velocity in this point. As you know the resulting speed is the hypotenuse of a right triangle with one side equal to the x-component and the other equal to the y- component.

 

[lwc729]

heres my solution...is it correct or close?

V_fy=V_iy+aΔt
=-2.8 m/s + -10 m/s(10.85 s)
V_fy=-111.3 m/s

Δx=(9.6 m/s)(10.85 s)
=104.16 m

Δy=30.392

tan^-1(30.392 m/104.16 m)
=16°

 

[nure]

The vertical distance is correct(i get 30.387 but i use a=9.81), but not the rest of it. Do not concentrate on the x-displacement. Concentrate on the speeds. You want the slope to be parallell with the velocity in the point 30 meters below the ski jump. Find the angle the VELOCITY makes with the horizontal/vertical. To find the velocity you need the velocity-components(speed) in two orthogonal directions(x and y).

From the top of the trajectory use the equation

V²=v₀²+2a(deltaY)

And solve for v. You then should get 24.42 m/s. And also, the acceleration will be positive on the way down.

 

[lwc729]

using your equation i came up with 24.49 m/s (due to my -10 m/s g)
combining the x and y, i used V_f=√(9.6 m/s)² + (24.49 m/s)²
=26.3 m/s

 

[lwc729]

from there i can't do the sin^-1 with two different units can i? 30.392 m for the y part in the triangle. 26.3 m/s for hypotenuse?

 

[lwc729]

i used 9.6 m/s from x and 26.4 m/s for hyp.
then cos^-1(9.6 m/s / 26.4 m/s)
=68.7°

closer??

 

[nure]

Correct! Hes velocity will make an angle 68.5 with the horizontal, and his resulting speed will be 26.2 m/s