Solving for x-Displacement of a Rolling Ball on an Inclined Track

[goodz]

Homework Statement

a track is inclined in such way that the velocity of the metal ball is 60 cm/s at 25 degree below the horizontal. mathematically determine the total x displacement of the metal ball.

weight of ball: 65grams
height from floor to the end of the track: 76.15cm
velocity of the ball is 60.cm/s at 25 degree

Homework Equations

$$\Delta$$y =Vo*t - .5 1/2gt^2
Vx = Vox + .5at^2

The Attempt at a Solution

I used the first equation to find time.
-76.15 = .5*-9.8t^2
t=.394s
deltaX = .6m/s*cos(-25).3942 + .5(-9.8).3942^2
i got delta X to be -54.7cm

is it suppose to be negative?

Could someone help me set this up right?

 

[rl.bhat]

Hi goodz, welcome to Pf.
In the y direction the initial velocity is not equal to zero.
And in the x direction there is no acceleration.

 

[goodz]

I think i got it.
first i needed to get the velocity using v^2 = VoSin(-25)^2 - 2(-9.8).7615
v = 3.87m/s
then to get time,
3.87 = .6Sin(-25)-(-9.8)t
t=.4207

 

[rl.bhat]

I think i got it.
first i needed to get the velocity using v^2 = VoSin(-25)^2 - 2(-9.8).7615
v = 3.87m/s
then to get time,
3.87 = .6Sin(-25)-(-9.8)t
t=.4207

If you take down as negative v, vi and g are all negative. So check the value of t.

 

[goodz]

are you saying .6m should be negative?

 

[rl.bhat]

are you saying .6m should be negative?

Allare positive or negative.
The equation should be
v = vosin(25) + gt.

 

[goodz]

3.87 = .6(sin25)+9.8t
t=.369s

x= voCos25t
x=.6cos25(.369)
x=.20m