(II) In the high jump, the kinetic energy of an athlete is tra...

Question

(II) In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10 m and cross the bar with a speed of 0.50 m/s?

Accepted Answer

Hello, let's go through this practice problem. What minimum initial speed is required for a basketball player to dunk the ball by elevating their center of mass by 2.11 m without mechanical aids during their jump while still maintaining a speed of 0.51 m per second at the peak of their jump. A 3.1 m per second. B 6.5 m per second. C 7.8 m per second or D 9.5 m per second. OK. So let's draw kind of a simplified picture of what's going on in this problem. So let's say this represents the ground and the basketball player wants to jump in such a way. I'm gonna draw a line representing the basketball player path as they jump. It's gonna jump in such a way that they come upwards up to a height of 2.11 meters. And even when they reach the peak of their, of their height, the peak of their motion at this apex, they're still moving at a speed, a horizontal speed of 0.51 m per second. Now, if we assume that gravity is the only force acting on the player during this motion, then we can solve this problem using the conservation of energy. Recall that the law of conservation of energy states that in the initial energy of the system must be equal to the final amount of energy in the system, assuming there are no act outside forces acting on this system. In the case of this problem, we're going to take the system of the basketball player and the earth, which means we're dealing with mechanical energy, which is the sum of kinetic energy and potential energy. So we can really write this equation as is that the initial kinetic energy of the basketball player recall that kinetic energy is one half MV squared. So one half multiplied by the mass of the player multiplied by the initial speed squared plus the initial potential energy, which has a formula recall that potential energy is equal to mass multiplied by the acceleration due to gravity multiplied by the height with respects to wherever our zero height is. So Y sub one is gonna be equal to the final kinetic energy plus the final potential energy. So again, the kinetic energy is one half MB squared only. Now we're going to use the subscripts two to represent the final speed plus the final potential energy. So mGy sub two representing the final height. Now a pretty easy way for us to simplify this equation is to define the figure, define our variables. So that the Y equals zero position is level with the ground, which means that our upper Y at the top of the player's motion is just going to be equal to 2.11 m, the distance between their current position and the ground. So if Y sub one is equal to zero, and that means that the entire mGy sub one term disappears and just goes to zero. Another way we can simplify is to notice that every single term carries a mass variable. So those can all cancel out. And we're left with the equation that one half V sub one squared is equal to one half V sub two squared plus GY sub two. We want to solve this problem for the initial speed required by the basketball player. So we're solving this equation for V sub one. So let's first get rid of the one half attached to the V sub one squared by multiplying all three equation all three terms in the equation or multiply both sides of the equation by two. So the left hand side becomes V sub one squared and the right hand side becomes V sub two squared plus two Gy sub two. And then we'll take the square root of both sides of the equation to get rid of that square on the V sub one. So the final equation we're left with is that V sub one is equal to the square root of the sub two squared plus two gy sub two. So all we gotta do is plug in the values that were given to us in the problem. So it's equal to the square root of V sub two squared. And we're told that V sub two, the height of the basketball player at the apex of their motion is equal to 0.51 m per second. And that's squared plus two multiplied by the G value 9.81 m per second squared multiplied by the height which is given as 2.11 m put all of this into a calculator. And we find that V sub one has a value of about 6.5 m per second. So our answer to the problem is 6.5 m per second which you'll notice corresponds to option B. So option B of this problem is our answer and that is it for this problem. I hope this video helped you out and please consider checking 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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(II) In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10 m and cross the bar with a speed of 0.50 m/s?

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