A basketball player grabbing a rebound jumps 60 cm vertically.(a) How much (total) time does the player spend in the top 25 cm of this jump?(b) How much (total) time does the player spend in the bottom 25 cm of this jump?

Question

agent0smith · Answer

Is this a physics or maths question?

anonymous · Answer

physics, should i move it ?

agent0smith · Answer

You don't have to I guess.

Use $$\Large h = \frac{ 1 }{ 2} g t^2$$where h=0.25m, this will give you HALF the time he spends in the top 25cm.

anonymous · Answer

so .25=1/2(-9.8)t^2

agent0smith · Answer

For the bottom 25cm:

you'll first have to find how long he spends in the air total, for 60cm (same calculation as above)

from that time, subtract the time he spends in the top 35cm (60-25 = 35) (same calculation as above)

anonymous · Answer

okay so first calculate 60=(1/2)(9.8)t^2=34.39

agent0smith · Answer

You'll need to use m, not cm.

anonymous · Answer

.60=(1/2)(9.8)t^2=3.429

agent0smith · Answer

t should be nowhere near 3 seconds... a person jumping 60cm will land again in under a second.

anonymous · Answer

is the acceleration not freefall?

agent0smith · Answer

Why wouldn't it be freefall?

anonymous · Answer

i just didnt know because i dont know why else i would be getting the number 3

agent0smith · Answer

Show your work because you're not doing something right...

anonymous · Answer

i put .60=(1/2)(9.8)(t^2)

agent0smith · Answer

@StephanieH Show how you solved for t.