A solid, uniform ball rolls without slipping up a hill, as shown in the figure (
ID: 1374078 • Letter: A
Question
A solid, uniform ball rolls without slipping up a hill, as shown in the figure (Figure 1) . At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V = 30.0m/s and H = 20.0m .
How far from the foot of the cliff does the ball land?
How fast is it moving just before it lands
Notice that when the ball lands, it has a larger translational speed than it had at the bottom of the hill. Does this mean that the ball somehow gained energy by going up the hill? Explain!
Explanation / Answer
7/10*vx^2 + 9.8*20 = 7/10*30^2
vx = 24.9 m/s
y= 1/2 * g*t^2
20 = 1/2*9.8*t^2
t = 2.02 s
a)
x = 24.9*2.02 = 50.3 m
b)
vy = 9.8*2.02 = 19.8 m/s
v = sqrt(vx^2 + vy^2) = sqrt(24.9 ^2 + 19.8^2)
v = 31.81 m/s
Notice that this is larger than the initial speed! The ball is spinning more slowly, so more energy is available for linear motion.