hang time: A basketball player, standing near the basket to grab a rebound jumps 76.0

Dorian Gray

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Jan 20, 2012
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Greetings Fellow Mathematicians,
I am having some difficulties with answering this question.

"A basketball player, standing near the basket to grab a rebound jumps 76.0 cm vertically. How much (total) time does the player spend
A) in the top 15.0 cm of this jump and
B) in the bottom 15.0 cm"


I THINK I have part A correct, but I am not absolutely sure. (I have the time as .175 but I put a note that it should be double (ie .350 s for up and down) )

As far as part B goes, I do not know how to address that part. I know that I will end up using one of the four formulas attached, but I cannot set up my equation(s).


Any and all help, comments, suggestions, ideas, tips are always welcomed and appreciated.

Thanks! jump.jpgequations.jpg
 
Greetings Fellow Mathematicians,
I am having some difficulties with answering this question.

"A basketball player, standing near the basket to grab a rebound jumps 76.0 cm vertically. How much (total) time does the player spend
A) in the top 15.0 cm of this jump and
B) in the bottom 15.0 cm"


I THINK I have part A correct, but I am not absolutely sure. (I have the time as .175 but I put a note that it should be double (ie .350 s for up and down) )

As far as part B goes, I do not know how to address that part. I know that I will end up using one of the four formulas attached, but I cannot set up my equation(s).


Any and all help, comments, suggestions, ideas, tips are always welcomed and appreciated.

Thanks! View attachment 2203View attachment 2204

For part b) - calculate -

i) how long does it take to travel 15 cm from ground ( against gravitational acceleration) ← corrected

ii) how long does it take to "fall" 76 cm (from 0 speed and positive gravitational acceleration)

iii) how long does it take to "fall" 51 (76 - 15) cm (from 0 speed and positive gravitational acceleration)

iv) total time i + ii - iii
 
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thanks!

Thank you S. Khan. Here is my work.


I am not sure if it is correct, but I tried my best to follow what you recommended.

Screen shot 2012-09-09 at 1.20.43 PM.jpg
 
Made a mistake in instruction for part (b)(i).

We need to have an initial velocity off of the ground - that cannot be zero.

So the jumper reaches a maximum height of 76 cm.

v2 = u2 + 2as

0 = u2 - 2*(9.8)*0.76

u = 3.86 m/sec

then time spent (t) to reach 15 cm

0.15 = 3.86*t - 4.9*t2

Quadratic equation → t = 0.041 sec

There are two solutions to 't'. Does that give you idea for solving this problem with a different strategy?
 
thank you!

Made a mistake in instruction for part (b)(i).

We need to have an initial velocity off of the ground - that cannot be zero.

So the jumper reaches a maximum height of 76 cm.

v2 = u2 + 2as

0 = u2 - 2*(9.8)*0.76

u = 3.86 m/sec

then time spent (t) to reach 15 cm

0.15 = 3.86*t - 4.9*t2

Quadratic equation → t = 0.041 sec

There are two solutions to 't'. Does that give you idea for solving this problem with a different strategy?

Greetings Subhotosh Khan,

Thank you very much for your responses. Just a few questions.

1. The t that you solved for: is that the time that we will add to ii and then subtract iii

2. For your question about the two "t"s and solving another way: Are you talking about having two equations and then using substitution?

Thank you very much again. This is my first ever physics class, and our professor doesn't give us much instruction on how to solve these until after the assignment is due.
 
Question:

For part b) - calculate -

i) how long does it take to travel 15 cm from ground ( against gravitational acceleration) ← corrected

ii) how long does it take to "fall" 76 cm (from 0 speed and positive gravitational acceleration)

iii) how long does it take to "fall" 51 (76 - 15) cm (from 0 speed and positive gravitational acceleration)

iv) total time i + ii - iii


Hello!

I was just wondering how you got the 51 in (iii), shouldn't be 61? That's the only doubt, otherwise this was a great explanation, I thank you for your time...
Thank you,

CJacob
 
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