A solid, uniform ball rolls without slipping up a hill as shown. Initially, at t

Question

A solid, uniform ball rolls without slipping up a hill as shown. Initially, at the bottom of the hill, the ball has a linear speed of 25.0 m/s. At the top of the 28.0 m high hill it launches off of a 30.0o slope into the air.
a) How far from the foot of the cliff does the ball land?
b) How fast is the ball moving just before it lands?
c) Compare your answer from b) with the initial linear speed of 25.0 m/s. Does you answer make sense? Explain.

A solid, uniform ball rolls without slipping up a hill as shown. Initially, at the bottom of the hill, the ball has a linear speed of 25.0 m/s. At the top of the 28.0 m high hill it launches off of a 30.0 degree slope into the air. a) How far from the foot of the cliff does the ball land? b) How fast is the ball moving just before it lands? c) Compare your answer from b) with the initial linear speed of 25.0 m/s. Does you answer make sense? Explain.

Explanation / Answer

Ebottom = KE + PE

KE = 1/2 m v12 + 1/2 I w2

I = 2/5 m r2

w = v / r

KE = 1/2m (v12 + 2/5 v12) = 7/10 m v12

PE = m g h ,[h = 0 ]

Etop = KE + PE

KE = 7/10 m v22

PE = m g h [h = 28 m ]

Ebottom = Etop

7/10 m v12 + 0 = 7/10 m v22 + m g h

v22 = v12 - 10/7 g h

v2 = sqrt ((25 m/s)2 - 10/7 * 9.8 * 28 m)

v2 = 15.26 m/s

h = 1/2 g t2

t = sqrt (2h/g) = sqrt [2 * 28 / 9.8]

t = 2.39 s

Distance traveled L = t * v2 = 2.39 * 15.26 = 36.45 m

----------------------------------------------------

vertical = sqrt (2 g h) = sqrt [2 * 9.8 * 28]

= 23.4 m/s

total velocity = sqrt (vhor2 + vvert2)

= sqrt (15.262 + 23.42)

= 27.9 m/s

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