Question help

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.
 

mezarashi

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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).
 
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
 

Fermat

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

Fermat

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Then it all falls out :smile:

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 ?
 
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
 

Fermat

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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.
 
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?
 

Fermat

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You used yf before as the height of the end of the slide. It would be inconsistent (and bad form :frown: ) 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.
 
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
 

Fermat

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The only acceleration involved is that of gravity.
 
so a would be 9.8?
 

Fermat

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yep, that's it.
 
could I use a projectile motion equation vy^2 = voy^2 - 2g(delta y)?
 
no wait Ineed to find the time so thats not the right equation
 

Fermat

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

Fermat

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Yes :smile:
 
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
 

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