Can someone answer this easy phsyics problem about vertical height

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Homework Help Overview

The problem involves calculating the vertical height attained by a basketball player during a dunk, given a hang time of 0.978 seconds and the acceleration due to gravity at 9.8 m/s².

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various attempts to calculate height using the formula H = 0.5gt² and question the reasoning behind using half the time squared for upward motion.

Discussion Status

There are multiple attempts to solve the problem, with participants providing different calculations and questioning the assumptions about time during ascent and descent. Some guidance has been offered regarding the relationship between time going up and time coming down.

Contextual Notes

Participants are navigating the constraints of providing help without directly solving the problem, emphasizing the need to show work for assistance.

Casey314stl
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Homework Statement

A basketball player achieves a hang time of 0.978 s in dunking the ball.
What vertical height will he attain? The
acceleration of gravity is 9.8 m/s^2

Answer in units of m

Homework Equations


?

The Attempt at a Solution

9.8/0.978^2= 10.245

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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Yes. You can.

We do not (can not) do homework for you. Show your work, we can help.
 
The attempt at a solution 9.8/0.978^2= 10.245
 
Casey314stl said:
The attempt at a solution 9.8/0.978^2= 10.245

Remember that H=.5gt^2

Retry and see what you get.
 
DaveC426913 said:
Yes. You can.

We do not (can not) do homework for you. Show your work, we can help.

I tried H=.5*g*t^2 and got 4.68677?
 
22.9653?
 
Casey314stl said:
I tried H=.5*g*t^2 and got 4.68677?

Remember that time going up = time coming down .
 
Casey314stl said:
22.9653?

No since t(up)=t(down)

then: H=.5g*(t/2)^2
 
mtayab1994 said:
Remember that time going up = time coming down .

what?
 
  • #10
mtayab1994 said:
No since t(up)=t(down)

then: H=.5g*(t/2)^2

Thanks a lot I got it but why is it half time^2?
 
  • #11
When you have the time going up = time going down, then you have: t(up)=t(down)=.978/2

I'm helping you too much.
 
  • #12
mtayab1994 said:
When you have the time going up = time going down, then you have: t(up)=t(down)=.978/2

I'm helping you too much.

I already got the the answer from the last response I was just wonder why it was half the time^2 if that was for every problem for this equation or just if an object is traveling up?
 
  • #13
Casey314stl said:
I already got the the answer from the last response I was just wonder why it was half the time^2 if that was for every problem for this equation or just if an object is traveling up?

Well since he is jumping upwards he only takes half of the time to get to his maximum height. Therefore the rest of the time is spent coming down. Hence no need for the time coming down.
 

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