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?

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

${V}_{f}^{2}$ = ${V}_{i}^{2}$ + 2a$\Delta \vec{x}$

${V}_{f}^{2}$ (final velocity) = ?

${V}_{i}^{2}$ (initial velocity) = 0 m/s

a (acceleration) = 9.8 m/${s}^{2}$

$\Delta \vec{x}$ (distance) = 11.8m

$\Rightarrow$ ${V}_{f}^{2}$ = 0 + (2 x 9.8 x 11.8) = 231.28

$\Rightarrow$ ${V}_{f}^{2}$ = 231.28m/s

Nov 15, 2015

$15.2 \text{m/s}$

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:

$m g h = \frac{1}{2} m {v}^{2}$

$\cancel{m} g h = \frac{1}{2} \cancel{m} {v}^{2}$

$\therefore {v}^{2} = 2 g h$

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

$v = 15.2 \text{m/s}$

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

Think of Galileo's famous experiment!