When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm. During the jump, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (c) In terms of this jumper's weight w, what force does the ground exert on him or her during the jump?

Question

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A kid's knees are initially on the ground, like in a kneeling position upon hopping straight from the state. The kid can reach a height of about 20 centimeters while making the hop. His body rises a distance of 16 centimeters from the knees up. Assume that this is true for his entire body as well. Solve for the force f that the ground exerts on the kid during the hop. Express this in terms of the kid's weight. W. So that's our end goal. Ultimately, we're trying to figure out what the value of F is the force that the ground exerts on the kid during the hop. And then we're also exact, asked to express this final answer in terms of weight. W Awesome. So with that in mind, let's note that all of our multiple choice answers are F the force is equal to something. So let's read them off to see what our final answer might be A is WB is 1.7 W C is two, W and D is 2.2 W. And note that it's a number multiplied by the value of W where the W is the weight awesome or the weight of the child to be exact. OK. So first off, let us note that the kid will jump with some initial velocity. But as soon as the kid attains the maximum height they can reach in their jump, the velocity will become zero. Thus, we can write and let's call this equation one, we can write that the final velocity VF squared is equal to the initial velocity squared minus two multiplied by the acceleration due to gravity multiplied by H the height. But we also need to note, let's make a little. And in this particular case, VF our final velocity so VF will equal zero. So VF our final velocity equals 0 m per second. OK. Which is for this particular case. Thus, we could write that rearranging the isolate in. So for V I, so V I squared is equal to two multiplied by G multiplied by H which is equal to two multiplied by our acceleration due to gravity, which we should recall to be the numerical value to be 9.81 m per second squared. And this is all multiplied by the height which is given to us as 20 centimeters, but we need to convert 20 centimeters to meters. So I've gone ahead and done this for you. But you can use dimensional analysis or quickly look it up, but it's 0.2 m. Thus, we could say that V I. So getting rid of the squared on V I, we can write that V I is equal to the square root of two, multiplied by 9.81 m per second squared multiplied by 0.2 m for the height. So when we plug that into our calculator, we will find that the initial velocity for this particular case is nine, so 1.98 meters per second. Fantastic. So now at this point, we need to consider when the kid is returning back to the ground and in that case, H will equal 0.16 m because once again, we had to convert centimeters to meters. So let's make a little note here. So now we're considering the case where H equals 0.16 m. So now we're focusing on this condition. So now in this case, we need to use equation one again. So we need to note that VF squared is equal to V I squared minus two multiplied by G multiplied by H. However, in this particular case, we need to note that the initial velocity for this case will be zero. So let's use our little blue pen again, make a little note here. So V I in this condition which we can draw another arrow. So we connect our stars here. So V I will equal 0 m per second. And we need to note that in this particular case, which once again, this is when the kid is returning back to the ground after jumping. So VF in this case will equal 1.98 m per second. Awesome. So now at this point, if we plug in all of our known variables back into equation one, we will get. And before we do that, let's take a note here that the acceler that in this case, we're not gonna use the value of gravity from equation one, we're gonna replace it with A Y to denote an acceleration because the acceleration will be different in this particular case. So with that in mind, we will write 9.8 so sorry, 1.98. So 1.98 m per second squared means that's our final velocity which is equal to 0 m per second squared minus two multiplied by. And now we're gonna plug in our value for a subscript Y for acceleration multiplied by 0.16 m. So at this moment, we do not know what the value of A Y is. So we need to rearrange this equation to isolate and sulfur A Y. So when we do that, we will find that A Y is equal to 1.98 m per second squared, all divided by two, multiplied by 0.16 m. Awesome. So let me plug that into our calculator. We will find that it's equal to 12.25 m per second squared. So 12.25 m per second squared, which is the standard units for acceleration. Now, at this point, we need to recall and apply Newton's first law. So let us recall that the sum of FY is equal to the mass multiplied by the acceleration of Y. So A Y which we can write as F minus W where W is the weight, so the force minus weight is equal to the mass multiplied by the acceleration of why? Which we can write as getting F by itself that F is equal to the mass multiplied by A Y plus W the weight awesome. And moving right along, we can say that if the force is equal to the weight multiplied by one plus A Y divided by G where G is the acceleration due to gravity. Thus, we can write when we plug in all of our known variables at this point that F is equal to the weight multiplied by one plus 12.25 m per second squared, all divided by 9.8. 1 means that the numerical value of the acceleration due to gravity. Therefore, when we plug that into our calculator, we will find that F is equal to 2.2487 W which rounding to two significant figures is equal to 2.2 W and that is our final answer is 2.2 multiplied by W. Hooray. We did it. Therefore, the force which the ground exerts is F equals 2.2. W where W denotes the value of the kid's weight. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter DF equals 2.2. W thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Verified video answer for a similar problem:

Key Concepts

Newton's Second Law of Motion

Gravitational Force

Kinematics of Vertical Motion