AP physics question...looking for the height of the waterslide:?

Question

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 the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.500 s after leaving the slide. Ignoring friction and air resistance, find the height H in the drawing.

_____m

link for the drawing:

http://www.webassign.net/CJ/06_45.gif

if anyone could explain how to do this then that would be great, thanks :)

Answer 1

For this question you must assume when the person leaves the end of the slide he is free falling with horizontal velocity.

V initial y = when he leaves the slide

V final y = when he hits the floor

y = height of fall from end of slide to floor

So since he is free falling:

V initial y = 0m/s

acceleration(a) = g

therefore we can use the equation a = (Vf- Vi) / t

so: g = V final y / t

g X t = V final y

9.8 X 0.5 = 4.9m/s =  V final y

so now we use equation:

Vfinaly ^2 = Vinitial y ^2 + 2ay

rearrange: Vfinal y^2 / 2g = y          [in the y direction a = g]

4.9^2/ (2 X 9.8) = 1.225 m = y        

We can also obtain velocity in the x direction (there's no acceleration according to Newton's law in the x direction):

avg velocity = d / t

= 5m / 0.5s

= 10m/s

So now we put everything together using energy:

Energy at the very top:

PE1 = mg(h-y)     [we use (h-y) because we consider velocity from height y ]          

KE1 = 0(because velocity is 0 at the top)

Energy at the bottom when person leaves slide:

PE2 = 0

KE2= 1/2mv^2

Conservation of energy tells us:

PE1 = KE2

mg(h-y) = 1/2m(v^2)      [mass cancels on both sides]

g(h-1.225) = 1/2(10^2)    [substitute]

h = 6.32m                          [rearrange and find h]

hmm... my answer seems to be different from theirs

well this is my best.

Answer 2

Hi friend!

I'm from Mexico, i don't speak english very well, but i will try it.

I' m very good in this, but i want to learn english so.. i think that Yahoo answers is a good way for do it.

Ok..

Let me see..

Your problem is easy

Look:

This problem we can do it using the energy's conservation and using the motion's equations.

Well, we must find the velocity in the moment when the boy leave the slide.

For this we can use the formula:

Vx = x / t

So:

Vx = 5 m / 0.5 seg

Vx = 10 m/s

Now, we use the energy's conservation

When the boy is in the high "H" he has potential energy (Ep), but when is just to leave the slide he has kinetic energy(Ek)

At one the energy's conservation,Ep and Ek will be iquals.

Remember:

Ep= m g H

and

Ek  = m v^2 /2

where

m = mass of the person

g = acceleration due to gravity = 9.8 m/sec^2 (is a constant)

H = height of the slide

V = velocity of person hitting the water = 10 m/sec.

Since:

Ep = Ek

mgH = mv^2/ 2

and since "m" appears on both sides of the equation, it will simply cancel out and the above simplifies to:

gH = v^2/2

H = v^2/ 2g

Using information:

H = (10 m/s)^2 / 2(9.8m/s^2)

H = 5.1020 meters ------------------->>> Answer

Please , check my mistakes and correct me.

See' ya

P.D. The second colaborator copy me check it

Answer 3

The velocity of the person as he hits the water = 5/0.5 = 10 m/sec.

Knowing this, apply the law of conservation of energy. i.e.,

Potential energy at top of slide = Kinetic energy as person hits the water

Potential energy = PE = mgH

Kinetic energy = (1/2)mV^2

where

m = mass of the person

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

H = height of the slide

V = velocity of person hitting the water = 10 m/sec.

Since PE = KE, then

mgH = (1/2)mV^2

and since "m" appears on both sides of the equation, it will simply cancel out and the above simplifies to

g(H) = (1/2)V^2

Solving for "H",

H = (1/2g)(V^2)

Substituting appropriate values,

H = (1/(2 * 9.8))(10^2)

H = 100/(2 * 9.8)

H = 5.10 meters