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?

2 Answers
Nov 15, 2015

231.28m/s

Explanation:

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 + 2aDeltavec 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

Nov 15, 2015

15.2"m/s"

Explanation:

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!