A 74 kg student starting from rest, slides down an 11.8 meter high water-slide. How fast is he going at the bottom of the slide?

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Answer 1 · 2015-11-15T13:06:57

231.28m/s

Sliding down, the major force acting in the student's weight.
Therefore, gravity is what is accelerating the student.
g = 9.8 m/#s^2#

Using the equation of motion

#V_f^2# = #V_i^2# + 2a#Deltavec x#

#V_f^2# (final velocity) = ?

#V_i^2# (initial velocity) = 0 m/s

a (acceleration) = 9.8 m/#s^2#

#Deltavec x# (distance) = 11.8m

#rArr# #V_f^2# = 0 + (2 x 9.8 x 11.8) = 231.28

#rArr# #V_f^2# = 231.28m/s

Answer 2 · 2015-11-15T18:40:41

#15.2"m/s"#

Applying the conservation of energy we can say that the potential energy at the top the slide will be converted to kinetic energy at the bottom:

#mgh=1/2mv^2#

#cancel(m)gh=1/2cancel(m)v^2#

#:.v^2=2gh#

#v=sqrt(2gh)=sqrt(2xx9.8xx11.8#

#v=15.2"m/s"#

As you can see, the mass of the student is irrelevant to the question.

Think of Galileo's famous experiment!