1. Sep 29, 2006

### meudiolava

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.

2. Sep 29, 2006

### Staff: Mentor

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.

3. Sep 29, 2006

### meudiolava

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

I used:

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.

4. Sep 29, 2006

### Staff: Mentor

What does the picture "shown below" show?

5. Sep 29, 2006

### Stevedye56

did you change everything into SI units?

6. Sep 29, 2006

### meudiolava

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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7. Oct 1, 2006

### Staff: Mentor

The distance is given as d = 83.7 cm = 0.837 m.

8. Oct 1, 2006

### Andrew Mason

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

AM