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 = 28.0 m/s and H = 24.0 m .

Part A

How far from the foot of the cliff does the ball land?

Part B

How fast is it moving just before it lands?

Thank you in advance for your help!

Answer 1

we need to find the velocity at the top of the cliff. We use conservation of energy.

E.bottom = KE + PE
KE = 1/2 m v1^2 + 1/2 I ω^2
I = 2/5 m r^2
ω = v/r
KE = 1/2m (v1^2 + 2/5 v1^2) = 7/10 m v1^2
v1= 28 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v2^2
PE = mgh
h = 24 m

E.bottom = E.top

7/10 m v1^2 + 0 = 7/10 m v2^2 + mg h
v2^2 = v1^2 -10/7gh
v2 = √((28m/s) - 10/7*9.81m/s^2 * 24m)
v2 = 27.99 m/s

1. time to fall
h = 1/2 g t^2
t = √(2h/g)

t=√2*24/9.81

t = 2.21 s

Distance traveled:
L = t*v2

L=2.21*27.99

L = 61.85 m ANSWER

2. v.vertical = √(2gh)

v.vertical=√2*9.81*24

v.vert = 21.69 m/s

total velocity = √ (v.hor^2 + v.vert^2)
= √ (27.99^2 + 21.69^2)
= 35.41 m/s