Part A A solid, uniform ball rolls without slipping up a hill, as shown in the f

Question

Part A 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 horizonlally, and then it goes over the vertical cliff How tar trom the toot ot the cit docs the ball land? Submit Part B Figure 1 of 1 How fast is it moving just betore it lands? 25.0 m/s m/s 28.0 m Submit Part c Nolice thal whehe balls lands, il has a grealer trans alional speed han when it was al the botlom of he hill. Dces lhis mean thal the ball someliow gained energy? Explainl

Explanation / Answer

You need to find the velocity at the top of the cliff. Use conservation of energy.

E.bottom = KE + PE
KE = 1/2 m v12 + 1/2 I 2
I = 2/5 m r2
= v/r
KE = 1/2m (v12 + 2/5 v12) = 7/10 m v12
v1= 25 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v22
PE = mgh
h = 28 m

E.bottom = E.top

7/10 m v12 + 0 = 7/10 m v22 + mg h
v22 = v12 -10/7gh
v2 = ((25m/s) - 10/7*9.81m/s2 * 28m)
v2 = 15.25 m/s

1. time to fall
h = 1/2 g t2
t = (2h/g)
t = 2.39 s

Distance traveled:
L = t*v2
L = 36.45 m

2. v.vertical = (2gh)
v.vert = 23.4 m/s

total velocity = (v.hor2 + v.vert2)
= (15.252 + 23.42)
= 27.9 m/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.