A basketball player jumps 76cm during a game. What time is

[Thor]

A basketball player jumps 76cm during a game.

A) What time is spent in the top 15cm of the jump? What time is spent in the bottom 15cm?

B) Define in words.

I have done part (B) already, but I don't understand part (A) at all. I tried using the free fall formula, s(t) = s0 + v0t - (1/2)gt^2, but it didn't work. I really need help. Thank you!

 

[stapel]

I tried using the free fall formula, s = s0 + v0t - 1/2g^2, but it didn't work.

Please reply showing how you used this formula (which should have worked).

Thank you.

Eliz.

 

[Thor]

I tried setting s as .76, because it's supposed to be in meters, and then I tried to find the time.

I kept getting answers that were WAY to high or way to low. And without initial velocity, there is nothing to suggest they even jump...I am really stumped on this equation, and I probably botched it on the test, since I couldn't ask questions about it till now. Any help is GREATLY appreciated.

 

[soroban]

Re: A basketball player jumps 76cm during a game. What time

Hello, Thor!

I see that you changed all measurements to meters . . . good!

A basketball player jumps 76cm during a game.

A) What time is spent in the top 15cm of the jump?
What time is spent in the bottom 15cm?

We must find the initial velocity, \(\displaystyle v_o\).

We know that the initial height is: \(\displaystyle s_o\,=\,0\),
. . so the height function is: \(\displaystyle \L\:s(t)\:=\:v_ot - 4.9t^2\)

When does he reach maximum height?
Set \(\displaystyle s'(t)\,=\,0\) and solve.
. . \(\displaystyle s'(t)\:=\:v_o\,-\,9.8t\:=\:0\;\;\Rightarrow\;\;\L t\,=\,\frac{v_o}{9.8}\) seconds.

The maximum height is 76 cm = 0.76 m.
. . We have: \(\displaystyle \L\:s\left(\frac{v_o}{9.8}\right)\;=\;v_o\left(\frac{v_o}{9.8}\right)\,-\,4.9\left(\frac{v_o}{9.8}\right)^2\;=\;0.76\)

Then we have: \(\displaystyle \L\:\frac{v_o}{19.6}\,=\,0.76\;\;\Rightarrow\;\;v_o\:=\:14.896\) m/sec.

Can you finish it now?

 

[Thor]

I believe so. Thanks for the help!