Question help

1. Nov 30, 2005

hannsparks

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As Figure 3 below shows, a 50 kg boy is observed to hit the water 7 meters from the end of the slide, as shown. The slide is frictionless. Also, ignore air resistance.
Find the speed, vB ,at which the boy leaves the slide, the initial height, H, of the boy, and the speed, vC, of the boy upon hitting the water.

I used the eqaution 1/2 mvi^2+mgyi=1/2 mvf^2 +mgyf to find the final velocity of the boy but I dont have the yi (initial height)

If you can tell me what to figure out first to get me going and how to do it I know I can figure out the rest.

2. Nov 30, 2005

mezarashi

The steps are:

1. Find the velocity of the boy as he leaves the horizontal end of the slide.
2. Find the time it takes for him to hit the water
3. Using that time, find the vertical distance he must have fallen from (i.e. how high is the slide above the water).

3. Nov 30, 2005

hannsparks

yi (the initial height of the boy) is unknown and I cant figure out the velocity of the boy as he leaves the slide at the horizontal without it, how would I go about figuring it out

4. Nov 30, 2005

Fermat

Have you given all the information ?

Do you know the height of the end of the slide above the water level ?

I've not been able to use the weight of the boy, which makes me wonder if all the info has been given.

5. Nov 30, 2005

hannsparks

yeah actually the height at the end of the slide is 2.4m ( i forgot to mention it)

6. Nov 30, 2005

Fermat

Then it all falls out

Have you got any work done, any equations ?

Oh yes, I see you've got something, but can you say where vf and yf apply to ? Is that the velocity at the end of the slide, or upon hitting the water ?

7. Nov 30, 2005

hannsparks

the vf that I was trying to calculate was the velocity of the boy as he comes off the slide and yf would be 2.4

8. Nov 30, 2005

Fermat

OK, that's one equation, in vf, yf and yi where yi and vf are the two unknowns.

Now analyse the motion of the boy after he has left the end of the slide (with horizontal velocity vf) as mezarashi detailed above.

9. Nov 30, 2005

hannsparks

so...
the equation for the two unknowns (yi and yf) will be
1/2(50(0)^2+ 50(9.8)yi=1/2(50)vf^2+(50)(9.8)(2.4)
490yi=25vf^2+1176

and to analyze the motion of the boy at the end of the slide
would you use the same equation but use yi as 2.4 and yf as zero becuase he is now in the water?

10. Nov 30, 2005

Fermat

You used yf before as the height of the end of the slide. It would be inconsistent (and bad form ) to use yf to denote another dimension.
Why not use yw (y at the water) as the height above the water, with yw = 0, of course. And keep yf = 2.4 m.

Also, you don't use the same eqn. That was the eqn for conservation of energy. You should use the kinematic eqns of motion for the travel of the boy during this period.

11. Nov 30, 2005

hannsparks

so when using the kinematic equation i would be delta y as 2.4 and solve for v but wouldn't I need to know the acceleration how owuld I do that

12. Nov 30, 2005

Fermat

The only acceleration involved is that of gravity.

13. Nov 30, 2005

hannsparks

so a would be 9.8?

14. Nov 30, 2005

Fermat

yep, that's it.

15. Nov 30, 2005

hannsparks

could I use a projectile motion equation vy^2 = voy^2 - 2g(delta y)?

16. Nov 30, 2005

hannsparks

no wait Ineed to find the time so thats not the right equation

17. Nov 30, 2005

Fermat

You know how high he is (2.4m). How long would it take for him to fall that distance under gravity ?
If his horizontal velocity is vf, how long would it take him to travel 7m (horizontally) ?
Equate the two times

18. Nov 30, 2005

hannsparks

deltay = voy t + 1/2 ay t^2 is it this one? where t=0.6998

19. Nov 30, 2005

Fermat

Yes

20. Nov 30, 2005

hannsparks

deltay = voy t + 1/2 ay t^2 y direction
deltax = vox t + 1/2 ax t^2 x direction

did i chose the right eqautions