Ball down a hill. What speed?

Question

Ball down a slope at an angle of 30 degrees

Ball rolls down a hill from a height of 10 metres at a slope of 30 degrees. What speed does it attain. I have put this question up before but have been frustrated when the answers, ignored the slope and assumed the ball falls vertically. Appreciate any help. Thanks

Answer 1

ou have many many scenarios.  i will go from the simple to the complex.

1.  Ball falling straight down. Assume vacuum near surface of Earth using the average gravitational force.  The ball will fall or accelerate downwards at 9.81 meters/s^2. (32 feet per second per second).  For formula to use is Velocity=ax.  where a is gee and x is distance.

2.  Ball falling straight down with air.  You must now examine the air resistance.

3.  Ball is SLIDING down a slope. assume no friction between ball and slope nor air resistance.  The gee is now at 1/2 because of the component of 30 degrees.

If it was 60 degrees, the value would be .866 gee.

4.  Ball down the slope with friction and no air.  Now you must add rotational inertia and acceleration.

5  Ball down the slope with all the friction.

4.  Ball is rolling down a slope with no air resistance

Answer 2

lets consider 2 cases

1) without consideration of slope

2) with consideration of slope

1) use formula d=0.5at^2

10=0.5(9.8)t^2

t= 1.428571429s

a=v/t

v=9.8(1.428571429)

v= 14 m/s

2) with slope acceleration due to gravity becomes 9.8cos (90-30)=4.9

now calculate velocity

10=0.5(4.9)t^2

t= 2.020306089s

a=v/t

v=4.9(2.020306089)

v= 9.9m/s

hope this helps

Answer 3

In this problem (based on the data given), you DO NOT need the slope of the incline. This piece of data is redundant.

Simply use conservation of energy,

Potential energy = Kinetic energy

PE = potential energy = mgh

Kinetic energy = (1/2)mV^2

where

m = mass of the ball

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

h = VERTICAL height of ball from reference line = 10 m (given)

V = velocity of ball at bottom of hill

Substituting appropriate values,

m(9.8)(10) = (1/2)(m)V^2

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

V^2 = 2(9.8)(10) = 196

V = 14 m/sec.

AGAIN, based on data given in this problem, the angle of the incline is a useless piece of data.