A person jumps out a fourth-story window 18.0 m above a firefi...

Question

A person jumps out a fourth-story window 18.0 m above a firefighter's safety net. The survivor stretches the net 1.0 m before coming to rest, Fig. 2–50. (a) What was the average deceleration experienced by the survivor when she was slowed to rest by the net? (b) What would you do to make it 'safer' (that is, to generate a smaller deceleration): would you stiffen or loosen the net? Explain.

Accepted Answer

Hello, let's go through this practice problem. A daredevil executes a high dive from a platform 30 m above the water's surface, an inflatable cushion is positioned in the water to ensure a secure landing. The cushion undergoes a 3 m expansion before gradually halting the daredevil's descent, determine one the mean deceleration encountered by the daredevil as they were brought to a halt by the inflatable cushion. And two, if the objective is to enhance the safety during the dive, by reducing the deceleration, should the inflatable cushion stiffness be increased or decreased. Option. A negative 98 m per second squared and two decreased. Option B one negative 49 m per second squared and two decreased option C one negative 98 m per second squared and two increased or D one negative 49 m per second squared and two increased. OK. So for part one of the problem, we're looking for the deceleration encountered by the daredevil as the inflatable cushion is slowing them down. So this is essentially a kinematics equation where we're assuming acceleration is constant because it's asking for the mean deceleration. And we're using the variables given the first thing we need to do is figure out what speed the daredevil actually has by the time they reach the cushion. So assuming the acceleration is constant, let's bring out the kinematics equations that we typically use for these problems. The first one stating that speed is equal to initial speed plus acceleration multiplied by time. The second one stating that displacement is equal to initial speed multiplied by time plus one half the acceleration multiplied by time squared. And the third one stating that final speed squared is equal to initial speed squared plus two multiplied by acceleration multiplied by displacement. The key with these types of problems is to choose the equation that includes the variable we're trying to solve for and doesn't include any other unknowns. The first part of what we're trying to do involves using the height of the daredevil falls. So the displacement to find the final speed and the only one of these three equations that includes both of those variables is the third equation the V squared equation. So according to the variables that we were given in the problem, the squared speed of the diver of the diver is equal to the square of its initial speed. And if he's jumping from rest, this is zero and zero squared to zero plus two multiplied by the acceleration which is the gravitational acceleration. So 9.81 m per second squared multiplied by the height he falls through which is 30 m as given in the problem, we put this into a calculator and we find that V squared is equal to 588.6 m per second squared. This is the value of these squared. Technically, these units are also squared, but we can take the square root of both sides to find that the speed itself is 24.261 m per second in a downward direction. So now that we know the divers speed when he reaches the cushion, now we'll take this speed and now we're going to treat it as an initial speed as we look at the divers deceleration as the cushion is accelerating him back up to basically slow him down. So we'll use the same equation, but now we're using it in a bit of a different way. We're now the V squared on the left is the square of the speed after the furthest extent of the cushions stretch. Don't forget to square it. And it's equal to the initial speed pres stretch, which is just our V and it squared again plus two multiplied by the acceleration multiplied by the extent of that stretch, which is just 3 m as shown in the diagram. So we're going to algebraically solve this equation for a defined that acceleration. So some pretty simple algebra here we subtract from both sides of the equation V not squared. So the stretch squared minus V not squared and divide by two multiplied by the displacement. So now we're gonna plug these values into our calculator, the ones that were given to us. So at, so the subs stretch is going to have a, a value of zero because at the furthest extent of the stretch, the diver is not moving, they are stretched out at the most and they're about to change directions and minus the square of the speed we found earlier of 24.261 m per second. Don't forget to square that divided by two multiplied by the extent of the stretch, which is 3 m. So if we put this into a calculator, then we find an acceleration of negative 98 m per second squared. And that is the answer to the first part of the problem. The second part of the problem asks how we can enhance safety during the dive. Should the inflatable cushion stiffness be increased or decreased? Now, the key here is to understand that it is safer and less damaging for the acceleration value to be slower. And as we can see from this equation, the acceleration will be smaller if the delta Y value which is in the denominator is going to be larger. And in order to make the stretch length longer, the cushion needs to be less stiff. So the answer to the second part of the problem is that the cushion stiffness should be decreased. So the answer to this problem is negative 98 m per second for part one and for part two decreased and option A is what agrees with that answer. So that is it for this problem? I hope this video helped you out and please check out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

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