[MAA_96]

If the ball moves from the top to the bottom with a speed of 14 m/s,neglecting the air resistance,how to find the height (h) ?

[dknt]

(Original post by MAA_96)
If the ball moves from the top to the bottom with a speed of 14 m/s,neglecting the air resistance,how to find the height (h) ?

I assume neglecting friction is in there too. The fact that the slide is oddly shaped doesn't matter. Consider the energy of the ball at the beginning and the end. Since there is no resistance there is no loss of energy i.e. Initial GPE = Final KE. You should find the mass terms cancel, giving you "h".

[MAA_96]

So ,mgh=mv^2 and we can cancel mass terms right ?

[dknt]

(Original post by MAA_96)
So ,mgh=mv^2 and we can cancel mass terms right ?

*(1/2)mv^2, and yes, the m will cancel.

[MAA_96]

But can we use the equation of motion to find the height ?

By using this equation





Vf= 14
Vi = 0
a = 9,8
d = ?

14^2 = 0 + 2*( 9,8) d
196 = 19,6 d
d = 10

[dknt]

(Original post by MAA_96)
But can we use the equation of motion to find the height ?

By using this equation





Vf= 14
Vj = 0
a = 9,8
d = ؟

14^2 = 0 + 2*( 9,8) d
196 = 19,6 d
d = 10

No, that's only for constant acceleration.

[MAA_96]

(Original post by dknt)
No, that's only for constant acceleration.

But ,why in the problem there is not constant acceleration ?

[dknt]

(Original post by MAA_96)
But ,why in the problem there is not constant acceleration ?

Would you go down that slide at the same rate as gravity would take you down? If one person went done the slide and another jumped off from the same height, they would not reach the ground at the same time. Even if the slide was straight diagonal, and was frictionless, the acceleration would be constant, but not 9.81m/s^2

[Stonebridge]

(Original post by MAA_96)
But ,why in the problem there is not constant acceleration ?

The steeper the slide, the greater the acceleration.
It has to be straight for the acceleration to be constant.
It will always be less than "g" except in the case where the slide is vertical.