Solving for Speed, Height of Boy on Water Slide

AI Thread Summary
The discussion focuses on calculating the speed and height of a boy on a water slide. The boy, weighing 50 kg, leaves the slide horizontally and hits the water 7 meters away, with the slide's height above the water being 2.4 meters. The initial speed as he leaves the slide is determined to be approximately 10 m/s, while the vertical speed upon hitting the water is calculated to be about 6.86 m/s. The initial height of the slide is found to be 7.5 meters using energy conservation principles. The conversation emphasizes the importance of using appropriate kinematic equations and understanding the motion in both horizontal and vertical directions.
hannsparks
Messages
31
Reaction score
0
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 don't 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.
 
Physics news on Phys.org
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 can't 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
 
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)
 
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
 
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?
 
  • #10
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.
 
  • #11
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
The only acceleration involved is that of gravity.
 
  • #13
so a would be 9.8?
 
  • #14
yep, that's it.
 
  • #15
could I use a projectile motion equation vy^2 = voy^2 - 2g(delta y)?
 
  • #16
no wait Ineed to find the time so that's not the right equation
 
  • #17
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
deltay = voy t + 1/2 ay t^2 is it this one? where t=0.6998
 
  • #19
Yes :smile:
 
  • #20
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 equations
 
  • #21
ok then the time for the horizontal would be 1.19522
 
  • #22
no i have to use the time it took him to hit the water to find the initial height I would use another kinematic equation but which one
 
  • #23
Errm, the time for the horizontal is exactly the same as for the vertical.
(which you got right)
The boy moves both vertically and horizontally in the same period of time.
 
  • #24
im stuck don't know what to do next
 
  • #25
t = 0.6998s
s = 7m
v = ?

but you can write,

v = s/t
v = 7/0.6998
v = 10 m/s (approx)
===============
 
  • #26
ok so the 0.6998 can be used to find the initial height
 
  • #27
ok so 7/0.6998 which equals 10m/s is the velocity as he leaves the slide
 
  • #28
ok 490yi=25(10^2)+1176
yi=7.5m
 
  • #29
ok last part the velocity right before he hits the water
vi=10
vf=?
a=9.8
delta x=7
delta y= 2.4
im i on the right track?
 
  • #30
I get 7.5m as well!
 
  • #31
vi = 10 m/s is correct.

But this is the vertical velocity (when he hits the water)
You already know the horizontal velocity, vf at the slide.

Use pythagoras to get the speed (do you know what I mean ?)
 
  • #32
a2 + b2 = c2?
 
  • #33
I assume you mean,

a² + b² = c²

or,

vf^2 + vi^2 = speed^2
 
  • #34
so it would be 10^2+ 0^2=speed^2
 
  • #35
no that doesn't make any sense because the speed would be the same as the final speed, so we have to calculate the vi some how
 
  • #36
would it make sense to solve for vi using the conservation of energy equation?
 
  • #37
Sorry, I've missed out a step.

I thought you had already worked out another velocity.

Ok, forget my earlier post about vi being a vertical velocity - that's wrong.

the 10 m/s, that is the horizontal velocity that the boy has when he leaves the slide. I think you called that Vb.

Ok, now we have to find the vertical velocity that the boy hits the water with.

He falls from a height h = 2.4 m ( the height of the slide above the water). He falls with a acceleration of g = 9.8 m/s². So,

v² - u² = 2as (with u,v,a,s as defined before)

v² - 0 = 2g*2.4
v = 6.858 m/s
===========

Then,
v² + Vb² = speed²
 
  • #38
wow! thanks fermat for all your patience with me I greatly appreciate it! I only discovered this website two days ago and I thank God I did you've been a tremendous help. thanks again
 
  • #39
Ciao :smile:
 

Similar threads

Calculate the speed of a boy at the end of the chute
Replies
8
Views
1K
If the height of the water slide is h = 3.2m and the person's initial
Replies
6
Views
6K
How Can Energy Methods Determine Ally's Initial Height From a Slide?
Replies
4
Views
2K
Sliding down a sphere of ice; when do you lose contact?
Replies
13
Views
4K
Horizontal impulse on a ball at rest on a plane (with friction)
Replies
14
Views
3K
At What Point Does the Boy Lose Contact with the Ice?
Replies
9
Views
3K
Height in conservation of energy problem
Replies
3
Views
3K
Boy Seated On Mound of Ice, Starts Sliding
Replies
4
Views
4K
What is the Solution for Finding Velocity and Distance on a Water Slide?
Replies
34
Views
4K
Girl on Slide: Solving for Normal Force, Energy Transfer, and Final Speed
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top