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PROBLEM 12-1 A 180-lb, 6-ft basketball player makes a jump shot Just before jumping, the basketball player crouches down 8 in If the player s leg muscles together apply an average force of 450 lb, how high will the basketball player jump
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To begin with, let s convert to metric units, rounding off to keep the numbers relatively simple 180 lb is 800 N The fact that the player is 6 ft tall is irrelevant (at least to us, perhaps not to the coach) Eight inches is 02 m And 450 lb of force is 2000 N In a real-life situation, the force applied would not be constant However, to keep things simple we can use the average force and assume a constant force (equal to the average force) during the time the basketball player is accelerating up from the bent-knee position Recall that force equals mass times acceleration (F 5 ma), so a constant force means a constant acceleration And a constant acceleration means that the velocity is increasing linearly (in a straight line) This last point is helpful, because it means that wherever we have a constant acceleration, instead of having to figure out the increasing velocity at each point in time, we can simply take the average of the initial and final velocity, and apply that average velocity over the same period of time and get the same mathematical result We break the problem up into the same steps as previously The first step is to determine maximum velocity at the end of step 1 Then using this as the starting velocity of step 2 and applying the downward acceleration of gravity, we calculate the distance traveled until the velocity is zero: This is the height of the jump Before we solve the problem, let s review and derive some basic formulas of motion By definition, speed or velocity is the distance traveled divided by time See Eq (12-1) This will be used to relate the velocity of the jump to the height of the jump (Throughout this section we will use speed and velocity interchangeably, Strictly speaking, velocity and acceleration are vectors, and speed is scalar But for our purposes we are only dealing with motion in the up and down direction Therefore to keep things simple, we will treat velocity and acceleration as scalars and use positive value for the up direction and negative value for the down direction) v 5 d /t (12-1)
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chapter 12 p h y s i o l o g i c a l a n d a n at o m i c a l B i o p h y s i c s
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The standard units are meters per second By definition, acceleration is the change in velocity divided by time a 5 (v2 2 v1)/t (12-2)
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Standard units are meters per second per second, or m/s2 When a force is applied to an object, the result is an acceleration that is equal to the force divided by the mass This is Newton s second law and is usually written as force equals mass times acceleration F 5 ma (12-3)
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The standard units of force are kg m/s2, also called a newton In step 1 of our problem we don t know the time over which the force is applied and we don t know the resulting velocity (that s what we re trying to calculate), so we re going to have to play with the above equations to get them into forms that contain what we do know We know the distance over which the force was applied (the 8-in, or 02-m, crouch), and we know the force and the mass which together can be used to calculate the acceleration Let s rearrange Eq (12-2) to put it in terms of what we want to calculate in step 1 v2 5 v1 1 at (12-4)
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This says that if an object is initially traveling at velocity v1 and we apply an acceleration a for t seconds, then the object will accelerate from velocity v1 to velocity v2 If the acceleration itself is changing, then we have to use integral calculus to calculate the average velocity during those t seconds But if the acceleration is constant, then the velocity change is linear (ie, a graph of velocity over time would be a straight line), so the average velocity is simply the average of the initial and final velocities vavg 5 (v1 1 v2) / 2 (12-5)
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Knowing the average velocity comes in handy because we don t know t (the amount of time the player is accelerating upward) Rearranging Eq (12-1), we can express the time in terms of the distance and the average velocity (which we do know) t 5 d/vavg (12-6)
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