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

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To determine the speed a basketball player must jump to reach a height of 83.7 cm, the player’s initial velocity can be calculated using the kinematic equation Vf^2 = Vi^2 + 2ad, where Vf is the final velocity at the peak (0 m/s), a is the acceleration due to gravity (-9.81 m/s²), and d is the jump height (0.837 m). The player bends his body down by 58.1 cm before jumping, but this does not affect the calculation for the jump height. The distance for the jump should be considered from the point of takeoff, which is the total height reached minus the height of the bend. The method used is correct, and the bending position does not influence the initial velocity calculation needed to achieve the desired height.
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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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