### Video Transcript

A hiker being chased by a bear runs
to the edge of a cliff and jumps off. The hiker is running horizontally
at 4.5 meters per second when he reaches the cliff edge. The hiker’s speed when he hits the
ground is 16 meters per second. What vertical downward distance did
the hiker fall? If the hiker stepped forward off
the cliff edge with negligible horizontal velocity, what speed would they hit the
ground at?

So we want to solve for the
distance the hiker falls. And we also wanna solve for the
speed the hiker would have if the hiker left the cliff edge with negligible
horizontal velocity. We can call this distance ℎ. And we can label the speed they hit
the ground at 𝑣 sub 𝑔. We’ll start by drawing a diagram of
the situation.

In the first scenario, our hiker is
running away from the bear at a speed of 4.5 meters per second in the horizontal
direction, before leaping off the edge of the cliff. We know that, under these
conditions, the hiker’s overall speed when he hits the ground is 16 meters per
second. And we want to use that information
to solve for the height of the cliff, ℎ.

If we approach this from an energy
perspective, we can say that the hiker’s energy, kinetic plus potential, just as he
leaves the edge of the cliff is equal to his energy just as he lands. Those are our initial and final
moments. In this system of the hiker and the
cliff, energy is conserved. So we can write that 𝐸 sub 𝑖 is
equal to 𝐸 sub 𝑓 and expand that expression to include kinetic and potential
energies.

If we set our zero value for height
at the base of the cliff, that means that when the hiker lands, he has no potential
energy due to gravity. So that term goes to zero. It turns out that’s the only term
we can eliminate because, initially, as the hiker leaves the cliff’s edge, he has
both gravitational potential energy and kinetic energy due to his motion.

Recalling the equation forms of
gravitational potential energy, 𝑚 times 𝑔 times ℎ, and kinetic energy, one-half
𝑚𝑣 squared, we can expand our energy balance equation. So it reads one-half the hiker’s
mass times 𝑣 sub 𝑥 squared, the horizontal speed as the hiker leaves the
cliff. Plus the hiker’s mass times 𝑔
times the height of the cliff is equal to one-half the hiker’s mass times 𝑣
squared. Where 𝑣 is the hiker’s overall
speed when he lands. We see that the mass of the hiker
appears in every term. So we can cancel it out. And since we want to solve for the
height of the cliff, ℎ, we rearrange this equation to do so. And we find that ℎ is equal to 𝑣
squared minus 𝑣𝑥 squared all over two 𝑔, where we treat the acceleration due to
gravity as exactly 9.8 meters per second squared.

Since we know 𝑣 and 𝑣 sub 𝑥,
we’re ready to plug in and solve for ℎ. When we do and enter these terms on
our calculator, we find that, to two significant figures, ℎ is 12 meters. That’s the height of the cliff from
which the hiker jumps.

Next, we picture a slightly
different scenario where, in this one, instead of the hiker running off the cliff’s
edge at speed, the hiker jumps off from rest. We want to solve for the speed of
the hiker when he hits the ground, 𝑣 sub 𝑔. And once again, our initial point
will be when the hiker jumps. And the final point will be when
the hiker lands.

Just like before, in this system,
energy is conserved. But unlike before, the initial
kinetic energy of our hiker is zero. In this case, our final potential
energy will again be zero since we let our height zero reference point be at the
ground level. Our equation, therefore, reduces to
PE sub 𝑖 equals KE sub 𝑓, where the potential energy is entirely due to
gravitation.

Referring once again to our
gravitational potential energy and kinetic energy expressions, we write that 𝑚𝑔ℎ
is equal to one-half 𝑚𝑣 sub 𝑔 squared. Once again, the mass of the hiker
cancels out from both sides. And rearranging to solve for 𝑣 sub
𝑔, we find it’s equal to the square root of two times 𝑔 times ℎ, the height of the
cliff. We know 𝑔 and we’ve solved for ℎ
in the earlier part of the problem. So we’re ready to plug in and solve
for 𝑣 sub 𝑔. When we enter these terms on our
calculator, to two significant figures, it’s 15 meters per second. That’s the hiker’s speed when he
reaches the ground after jumping from rest.