Video: Using the Relationship between Speed, Distance, and Time to Find the Distance

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial vertically downward velocity of 1.40 m/s and observes that it takes 1.80 s to reach the water. How high above the water was the preserver released?

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Video Transcript

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial vertically downward velocity of 1.40 meters per second and observes that it takes 1.80 seconds to reach the water. How high above the water was the preserver released?

As we work through this problem, we’ll assume that the acceleration due to gravity, 𝑔, is exactly 9.8 meters per second squared. Let’s highlight some of the important information we’ve been given. We’re told that the preserver is thrown downward with an initial speed of 1.40 meters per second. We’ll call that 𝑣 sub 𝑖. We’re then told that the life preserver takes 1.80 seconds to reach the water. We’ll call that value 𝑡. We want to solve for the height above the surface of the water that the preserver was released. We’ll call that height ℎ.

Let’s begin our solution by drawing a diagram of the scenario. The helicopter is hovering over the water and one of the rescuers throws the life preserver downward with an initial speed 𝑣 sub 𝑖. We choose to let motion downward be in the positive direction. We want to figure out the height, ℎ, from which the preserver was thrown. As soon as the preserver was released, it was subject to only one vertical force, the force of gravity. Therefore, its acceleration was constant, so we can use a kinematic equation to investigate this problem.

Looking over these equations, we’d like to find a match between the information we’ve been given and what we’re looking for, height ℎ. The third kinematic equation from the top is a good match. In our case, 𝑑, the displacement, is the height ℎ that we’re solving for. 𝑣 sub zero is 𝑣 sub 𝑖, the initial speed of the preserver. The time value 𝑡 is given to us as 1.8 zero seconds, and the acceleration 𝑎 is 𝑔, 9.8 meters per second squared.

When we enter in all these values, we see that both 𝑣 sub 𝑖 and 𝑔, the acceleration due to gravity, are positive because they act in the downward direction. To three significant figures, we calculate ℎ to be 18.4 meters. That’s how far above the water the life preserver was released.

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