A daring 510-N swimmer dives off a cliff with a running horizonta...

Question

A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?

Accepted Answer

Hi, everybody. Let's get started. So a playground is located 3.5 m from the base of a cliff. This 3.5 m wide zone from the cliffs bases used by cheering Spectators. A player at the top of the cliff launches a ball with a force of 4.50 newtons horizontally. What must be the minimum speed of the ball as it leaves the top of the cliff so that it misses the spectator area at the base of the cliff. The spectator area is 3.5 m wide and 12.0 m below the top of the cliff. So We need to figure out our V and were given a few variables here. We have, we need to figure out a horizontal and vertical displacement. So because we're only looking at a minimum speed, we need to, we only need to figure out The speed required to kick it at least 3.5 m so that it misses our spectator area. And then since the ball is 12 m above the ground, gravity is going to pull it down negative 12 m. So that will be our displacement in the Y axis. Now we also have a V0 and I'm gonna call this V zero X. So our initial speed in the X direction that is our own known, that's what we want to figure out. But we also have an initial speed in the Y axis of zero m per second. That's because when the ball is kicked, of course, because it's being kicked horizontally, there's no force acting on it initially other than gravity. So the initial speed is going to be zero and then it's going to be accelerating downward as soon as it gets kicked. Now, how does that help us? Well, we can use our kid a matics equations here for the displacement in the Y axis. And because we're given the zero Y if you recall our equation here, the displacement in the Y is equal to the V zero Y times time plus one half the acceleration times times square. So here we know everything except for time. And we're going to need that because in order to figure out a minimum speed we need and we're given the displacement, we also need to find the time so that we can multiply those together. So we're going to substitute Are negative 12 m displacement. We have this whole section of course, is going to equal out to zero because it's zero times time. And then we're going to add that to one half. Our acceleration here is going to be G the acceleration due to gravity. So 9.8 m per meters per second squared times T squared, which is our unknown. So if we rearrange this problem, we get T squared equals negative 12.0 m minus one half divided By 9.8 m/s squared. And then because we just want time, we're going to take the square root to cancel that out. And you get this silly little equation here. And if you calculate that out, you should get a time of 1. seconds. Now, what can we do with that? Now, we can take our time and figure out How our displacement in the X, how much speed that that will give us. So our displacement in the X is going to be our V in the X axis times time. So We know our displacement needs to be at least 3.5 m and we need to find our initial speed. So we're gonna leave that as a variable. And now we know that the ball is going to travel 1.56 seconds before it hits the ground because of what we learned from our wide displacement. So if we divide our 3.5 m by our time, we will get our V zero and the X axis which is going to be 2.24 meters per second. So if we go check our answer choices, You can see answer choice. A 2.24 m/s is going to be the correct answer. Thank you.

A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?

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