What is the ball’s speed as it emerges from the pipe onto the green?

A miniature golf course has a hole in which the fairway is 3.6 m above the green. If you hit the ball into the middle hole in a row of three, the ball will be directed to the green by a connecting pipe. Suppose the ball falls down most of the length of the pipe and slides the rest of the way without any loss of energy to friction. What is the ball’s speed as it emerges from the pipe onto the green?

1 Answer
NJ
Apr 3, 2018

#=> v = 8.5# #"ms"^-1#

Explanation:

The moment the ball enters the pipe, it will possess some amount of potential energy.

#=>E_p = mgh#

where #m# is the mass of the ball, #g# is the gravitational acceleration constant, and #h# is the height.

The ball loses no energy while it travels through the pipe. When it reaches the bottom, all of its potential energy will have converted into kinetic energy. Kinetic energy is defined as:

#=>E_k = 1/2 m v^2#

where #m# is the mass of the ball and #v# is the speed of the ball.

Since energy is conserved, the initial energy must equal the final energy.

#=>E_p = E_k#

#=> mgh = 1/2 mv^2#

#=> cancelm gh = 1/2 cancel m v^2#

#=> gh = 1/2 v^2#

Solving for #v#:

#=> v = +-sqrt(2gh)#

Since we only care about the scalar speed of the object, just take the positive result:

#=> color(green)(v = sqrt(2gh))#

Now we use the height of #h = 3.6# #"m"# and for convenience I'll use #g = 10 # #"ms"^-2# (you should use whatever constant #g# you're expected to use, such as #9.8 # #"ms"^-2#).

#=> v = sqrt(2*10*3.6) = 8.5# #"ms"^-1#

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