How Fast Must a Basketball Player Jump to Reach 83.7 cm?

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Homework Help Overview

The problem involves a basketball player who needs to determine the speed required to jump to a height of 83.7 cm. The player bends his legs, dropping his upper body by 58.1 cm before jumping, which adds complexity to the situation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for a baseline height from which the player jumps and the application of kinematic equations to find the initial velocity. There are questions about the relevance of the player's bent position and the correct interpretation of the distance involved in the jump.

Discussion Status

Some participants have attempted calculations using kinematic equations but report that their answers are incorrect. There is ongoing exploration of the assumptions regarding the player's position and the distances involved, with suggestions for clarifying the initial conditions.

Contextual Notes

Participants are considering the implications of the player's bent position and whether it affects the calculations. There is also a focus on ensuring all measurements are in SI units.

meudiolava
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A 93.7 kg basketball player can leap straight up in the air to a height of 83.7 cm, as shown below. The player bends his legs until the upper part of his body is dropped by 58.1 cm, then he begins his jump. With what speed must the player leave the ground to reach a height of 83.7 cm?

I'm really stumped on how to do this question. I tried calculating the velocity he has when he reaches the ground, but that isn't right.

I don't really know what to do with the distance he drops his body by either.

A little help please?
 
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Does the question state how high the player can reach when standing on their toes? That's the baseline height that is needed, IMO. That is the position the player is in as they "leave the ground", and then you just use the standard kinematic equations to calculate the Vo needed to get whatever the delta-y is up to 83.7cm.
 
That's what i did, but the answer is wrong.

I used:

Vf^2 = Vi^2 + 2ad

I made:
a= -9.81 (because its in the negative direction)
d=8.31 m
and Vf 0 (once he reaches the top he has no more velocity). Then I isolated for Vi.

I think I have to do something with the fact that he's bent over a bit, but I have no idea what. I also don't understand why my method isn't working.

Any suggestions would be much appreciated.
 
What does the picture "shown below" show?
 
did you change everything into SI units?
 
Here is the picture. I hope this works!

And yes, I changed it to SI units. Thanks for the suggestion anyways!
 

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meudiolava said:
That's what i did, but the answer is wrong.

I used:

Vf^2 = Vi^2 + 2ad

I made:
a= -9.81 (because its in the negative direction)
d=8.31 m
The distance is given as d = 83.7 cm = 0.837 m.
 
meudiolava said:
I think I have to do something with the fact that he's bent over a bit, but I have no idea what. I also don't understand why my method isn't working.

Any suggestions would be much appreciated.
Your method is correct. The fact that he is bent over appears to be irrelevant to this part of the question.

AM
 

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