Calculate the speed of a boy at the end of the chute

[Alexanddros81]

Homework Statement

14.11 A boy slides down a water chute, starting from rest a A. Neglecting friction
determine his speed (a) at the end B of the chute; and (b) on entering the water at C.

Homework Equations

The Attempt at a Solution

for (a):

for (b):

Do I have to treat the boy at B as a projectile?

 

[kuruman]

Do I have to treat the boy at B as a projectile?

The answer is "No" if all you have to do is find the speed at C. You would if you had to find the time he sails through the air.

 

[Alexanddros81]

Am I using again principle of work and energy for (b)?
Why the 30deg angle is given at point B? Is it used in the calculations?

 

[kuruman]

Am I using again principle of work and energy for (b)?

Yes.

Why the 30deg angle is given at point B? Is it used in the calculations?

Probably to show you that you can use energy considerations or projectile motion and get the same answer either way. Try it and see for yourself.

 

[lekh2003]

for (b):

Do I have to treat the boy at B as a projectile?

Yes, you must treat the boy as a projectile. The boy will have a horizontal velocity you calculated above and zero vertical velocity. Zero horizontal acceleration and gravity pulling downwards.

You could also do this with the work energy principle of course, if you wanted. As explained by kuruman

 

[kuruman]

The boy will have a horizontal velocity you calculated above and zero vertical velocity.

The launch speed as calculated by OP is 12.52 m/s. The velocity is directed at 30^o above the horizontal and has both a vertical and a horizontal component.

 

[lekh2003]

The launch speed as calculated by OP is 12.52 m/s. The velocity is directed at 30^o above the horizontal and has both a vertical and a horizontal component.

Oh yes, sorry. I made a mistake.

 

[Alexanddros81]

So here is my solution using work and energy principle:

Is this correct?
Also I used projectile motion and I got $v_C=13.92 m/s$. I guess it is the precission in the decimal places of the numbers used for the calculations.

 

[kuruman]

I guess it is the precission in the decimal places of the numbers used for the calculations.

I am sure it is round off errors. Also note that for part (b) you could have conserved energy directly from A to C and used $U_{A \rightarrow C}=W(10~m)$.