What Are the Correct Launch Parameters for a Hollywood Daredevil's Canyon Jump?

[lizbette]

physics daredevil, wooo!

Homework Statement

A Hollywood daredevil plans to jump the canyon shown here http://img39.imageshack.us/img39/9406/jonesf336.gif on a motorcycle. If he desires a 3.0-second flight time, what is a) the correct angle for his launch ramp, b) his correct launch speed, c) the correct angle for his landing ramp, and d) his predicted landing speed? (Neglect air resistance)

Also, in the picture, the ramp length is 60m. I'm not sure if that makes a difference.

Homework Equations

yf=y0+voyt-1/2 gt^2
vf^2=voy^2-2gy
vy=voysintheta

The Attempt at a Solution

so since this flight is all due to velocity in the y direction, we shouldn't consider vx because it's zero all the way, right?
so we're left with voy. we know that yo is 15m, yf is 0m, and t is 3.0s
so 0=15+voy(3)-4.9(9)
voy=9.7m/s
so then you plug that into equation #2:
I'm not completely sure if in this case y=15m
vf^2=9.7^2-2(9.8)(15)
but then I get vf^2=-199.91
and that's not possible.
vf=14.14m/s?
and if vf really is 14.14m/s, how do I find the angle?

 

[mgb_phys]

I think you mean the X acceleration is zero!
If X velocity is zero you aren't going to go very far!

 

[nickjer]

A couple of things to note:

If v_x = 0, then he wouldn't move horizontally at all. Which would make the jump impossible. But whatever his initial v_x is, it will remain constant throughout the entire jump, since there is no acceleration in the horizontal direction.

Also, in equation #2, the "y" should actually be the "change in position" ($\Delta y$). So that would be $\Delta y = y_f - y_0 = 0 - 15 = -15$. So you were off by a negative sign for that term.

EDIT: Just realized you posted this twice. Please refrain from posting a question twice in the forums. It tends to upset the moderators.

<Moderator's Note: Threads merged.>

 

[Neesha]

i don't think you have enough info because to calculate the vf you need the horizontal velocity vx.you use Pythagoras to find the vf. vx^2 + voy^2 gives you vf and you find the angle using tan$$\theta$$ = voy/vx
i got the same voy as you but i don't think that how u calc vf. i was taught this way but i need the horizontal displacement to find vx using the formula v=s/t

 

[mgb_phys]

You do have enough info, if they are going to be in the air for 3,0sec and you know they path (a parabola) there is only one speed/angle they can start at.

hint: imagine them dropping a object from the top of their jump - they fall down at exactly the same vertical rate

 

[Delta2]

I think too that u need the horizontal travel distance x, so u can calculate $$v_{0x}=\frac{x}{3.0}$$ and then calculate the angles $$tan(launch)=\frac{v_{0y}}{v_{0x}}$$ and $$tan(landing)=\frac{v_{0y}-3g}{v_{0x}}$$ and for velocities would be

$$v_{launch}^2=v_{0x}^2+v_{0y}^2, v_{landing}^2=v_{0x}^2+(v_{0y}-3g)^2$$

 

[thrill3rnit3]

You'll need the horizontal distance. With your current information, you can deduce Voy without any problems. That's as far as you can go though, I think.

You can spend 3 seconds going up, or 3 seconds going forward, that's why you'll need the horizontal displacement.