Video: Kinematics of Conversion between Gravitational Potential Energy and Kinetic Energy

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 m/s when he reaches the cliff edge. The hiker’s speed when he hits the ground is 16 m/s. 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?

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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.

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