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 s in the air (their "hang time"). Treat Kobe as a particle and let ymax 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 Note from Asker: I am generally confused on how to manipulate the formulas, so if you could show every step that would be great, Thank You. Part A Part complete To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may 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 s in the air (their "hang time"). Treat Kobe as a particle and let ymax 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 Note from Asker: I am generally confused on how to manipulate the formulas, so if you could show every step that would be great, Thank You. Part A Part complete To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may 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 s in the air (their "hang time"). Treat Kobe as a particle and let ymax 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 Note from Asker: I am generally confused on how to manipulate the formulas, so if you could show every step that would be great, Thank You. Part A Part complete To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may 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 s in the air (their "hang time"). Treat Kobe as a particle and let ymax 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
Note from Asker: I am generally confused on how to manipulate the formulas, so if you could show every step that would be great, Thank You.
Part A
Part complete
To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.
_________
Study of objects in motion.
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