Calculating Time for Ball Rolling Off a Roof

[Vanessa23]

Homework Statement

a roof makes an angle of 30deg with the horizontal. A ball rools off the edge at a speed of 5.0 m/s. The distance to the ground is 7meters. How long is the ball in the air?

Homework Equations

0=volt-1/2gt^2+y
t=(vo+/- sqrt(vo^2-(4*gy)))/g

The Attempt at a Solution

0=.5*(-9.81)(sin(30))*t^2 + 5t + 7
from quadratic equation I get t=.468 but it should be .97 seconds.
I tried using cos(30) instead and I got 2.0 seconds. Please point out what I am doing wrong. Thanks!

 

[BatmanACC]

Hey there. You are going about it the wrong way. You need to decompose your vector into vertical and horizontal components.

After decomposing you should find your vertical component is 2.5m/s. This is your V1. Now we analyze vertically.

We use the equation D= V1 * T + .5(A)T^2

We have D, V1, and A. Input those values, re-arrange and you get a quadratic equation. Put that into the formula and you'll get .969s or a negative value. We discard the negative because time cannot be negative.

Hope that helps :)

 

[Vanessa23]

You say vertical component twice. I am assuming the 2.5 m/s is the horizontal, but how did you get that? And do I plug in sin(30)*9.81 for the acceleration for the y component in the equation we both used?

 

[BatmanACC]

You say vertical component twice. I am assuming the 2.5 m/s is the horizontal, but how did you get that? And do I plug in sin(30)*9.81 for the acceleration for the y component in the equation we both used?

No 2.5 is the vertical component.

Use sin/cos/tan. You have 5m/s vector at 30 degrees to the horizontal. Thus decomposing it means. 5sin30 = vertical.

Use that value (2.5) in the equation D= v1 * Time + 1.5(a)t^2

It'll come out to your answer. I already checked it twice for you.

 

[Vanessa23]

So I must be doing something else wrong because I am not getting .97 seconds. Here is what I am doing:

(-2.5)-sqrt(2.5^2-(4*7*.5*-9.81)) all divided by 9.81... equals 1.48s

 

[BatmanACC]

So I must be doing something else wrong because I am not getting .97 seconds. Here is what I am doing:

(-2.5)-sqrt(2.5^2-(4*7*.5*-9.81)) all divided by 9.81... equals 1.48s

7m = 2.5 x T + 4.9T^2

4.5T^2 + 2.5T - 7 = )

-2.5 square root ( 2.5^2 -(4)(4.9)(-7))
/
9.8

-2.5 +or- 12
/
9.8

X1= .969
X2= Negative

 

[Vanessa23]

Thanks for sticking with me! I'll know now that the negative goes with the distance (c) rather than the acceleration (a).

 

[Vanessa23]

So let's say that the ball started at a height of 12.5 meters and rolled to the edge of the roof which is at 7 meters. Using a^2+b^2=c^2 we know it rolled a distance of 8.13 meters on the roof. To find the time it rolled on the roof, you would use
acceleration=g*sin(theta)-g*(coeff. of friction)*cos(theta)
to get acceleration and then can you plug that into
V=Vo+at
if you have found the final velocity (V) from using PE=KE+Ffriction? Or do you have to solve another quadratic?

 

[BatmanACC]

So let's say that the ball started at a height of 12.5 meters and rolled to the edge of the roof which is at 7 meters. Using a^2+b^2=c^2 we know it rolled a distance of 8.13 meters on the roof. To find the time it rolled on the roof, you would use
acceleration=g*sin(theta)-g*(coeff. of friction)*cos(theta)
to get acceleration and then can you plug that into
V=Vo+at
if you have found the final velocity (V) from using PE=KE+Ffriction? Or do you have to solve another quadratic?

Hold the phone. You know the starting height and the height at the edge of the roof. The difference between them is your vertical component of your triangle. The difference is 5m (12-7). We have a right angled triangle with a 30 degree angle to the horizontal. Meaning sin30=5/h and h=5/sin30. Your hypotenuse is 10m not 8.13.

Assuming the ball wasn't pushed off the roof acceleration is still 9.8m/s (gravity). If it was pushed however we would need more information. Something like the time it took from the top of the roof to the bottom of the roof.

So v2 simply is (v2-v1)/T=a

 

[Vanessa23]

Sorry, didn't give all the info:
ball starts at height 12.5 to roll down roof at angle 40deg. the adjacent length (horizontal distance) to the edge of the roof is 6m. So the triangle is 5.5x6x8.14m.

so to find the time I did the following:
The roof makes a 40deg angle with the horizontal and coeff of friction is .388.

Potential E=KE+Ffriction
mg*distance*sin(theta)=.5mv^2+(coeff of friction)*mgcos(theta)*distance down roof
acceleration=g*sin(theta)-g*(coeff. of friction)*cos(theta)
V=Vo+at
So...
8.14*g*sin(40)=.5v^2+.388*g*cos(40)*(8.14)
v=9.84m/s

a=g*sin40)-g*.388cos(40)
a=3.39m/s^2

9.84=3.39t
t=2.90 seconds

 

[BatmanACC]

Sorry, didn't give all the info:
ball starts at height 12.5 to roll down roof at angle 40deg. the adjacent length (horizontal distance) to the edge of the roof is 6m. So the triangle is 5.5x6x8.14m.

so to find the time I did the following:
The roof makes a 40deg angle with the horizontal and coeff of friction is .388.

Potential E=KE+Ffriction
mg*distance*sin(theta)=.5mv^2+(coeff of friction)*mgcos(theta)*distance down roof
acceleration=g*sin(theta)-g*(coeff. of friction)*cos(theta)
V=Vo+at
So...
8.14*g*sin(40)=.5v^2+.388*g*cos(40)*(8.14)
v=9.84m/s

a=g*sin40)-g*.388cos(40)
a=3.39m/s^2

9.84=3.39t
t=2.90 seconds

In the equation that you stated it seems that you require Mass. I didn't see any listed. Perhaps you had the mass cancel out but I don't see that. Correct me if I'm wrong.

But if you have the mass (assuming you just didn't post it) Than it would be easier to equate PE and KE

Find the potential energy at the top VIA MGH. Once that is found equate that value to KE (.5mv^2) - Ff - MGH (new potential energy at new height).

You find your FF via UkFn. That seems much simpler. Get back to me on the mass thing though

 

[Vanessa23]

the mass cancels out, we don't know the mass.

 

[Vanessa23]

So, you mean I should have done:
g*12.5=.5*v^2-.388*g*cos(40)*8.14-g*7
v=17.1 m/s

so then
17.1=3.39t
t=5.05 sec