In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their “hang time”). Treat the athlete as a particle and let y max be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above y max /2 to the time it takes him to go from the floor to that height. Ignore air resistance.
In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their “hang time”). Treat the athlete as a particle and let y max be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above y max /2 to the time it takes him to go from the floor to that height. Ignore air resistance.
In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their “hang time”). Treat the athlete as a particle and let ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height. Ignore air resistance.
In the vertical jump, an Kobe Bryant starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 ss in the air (their "hang time"). Treat Kobe as a particle and let ymaxymax be his maximum height above the floor. Note: this isn't the entire story since Kobe can twist and curl up in the air, but then we can no longer treat him as a particle.
Hint: Find v0 to reach y_max in terms of g and y_max and recall the velocity at y_max is zero. Then find v1 to reach y_max/2 with the same kinematic equation. The time to reach y_max is obtained from v0=g (t), and the time to reach y_max/2 is given by v1-v0= -g(t1). Now, t1 is the time to reach y_max/2, and the quantity t-t1 is the time to go from y_max/2 to y_max. You want the ratio of (t-t1)/t1
To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore…
High-speed motion pictures (3500 frames/second) of a
jumping 230 μg flea yielded the data to plot the flea's
acceleration as a function of time as shown in the figure (
Figure 1). (See "The Flying Leap of the Flea," by M.
Rothschild et al. in the November 1973 Scientific
American.) This flea was about 2 mm long and jumped at
a nearly vertical takeoff angle. Use the measurements
shown on the graph to answer the questions.
Figure
alg
150
100
50
0
0
0.5
1.0
1.5
Part E
Use the graph to find the flea's maximum speed.
Express your answer in meters per second to two significant figures.
IVE] ΑΣΦ 3
v=1.6
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The figure here shows the speed v versus height y of a ball tossed directly upward, along a y axis. Distance d is 0.37 m. The speed at height yA is vA. The speed at height yB is vA/3. What is speed vA?
Chapter 2 Solutions
University Physics with Modern Physics, Volume 1 (Chs. 1-20) and Mastering Physics with Pearson eText & ValuePack Access Card (14th Edition)
Sears And Zemansky's University Physics With Modern Physics
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