# Challenging problem (free fall)

1. Jan 25, 2008

### Cathartics

[SOLVED] Challenging problem (free fall)

I am stuck with couple of problems and i keep
getting the wrong answer. Perhaps my approach is wrong. Below
are the questions and they way i approached the problems.

A basketball player grabbing a rebound jumps 80 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?

Ok so here is what i did. I took Vf=0 at the top. Vo= unknown,
a = -9.8, Y = .25m so i used the formula Y= ((vf^2 - Vo) / 2a)
And i got Vo = 2.21ms Now I used Vo in the equation Y = (vf+Vo)t
To find out T= .22s now the time going up is equal the time
coming down which is 2t = .45s
But it tells me that my answer is wrong.And how should i
approach the problem (b)???
-------------------------------------------------------
A lead ball is dropped into a lake from a diving board 5.06 m
above the water. It hits the water with a certain velocity and
then sinks to the bottom with this same constant velocity. It
reaches the bottom 4.92 s after it is dropped. (Assume the
positive direction is upward.)

(a) How deep is the lake?
(b) What is the average velocity of the ball?
(c) Suppose that all the water is drained from the lake. The
ball is now thrown from the diving board so that it again
reaches the bottom in 4.92 s. What is the initial velocity of
the ball?

(a)I put the problem in two parts. When the ball is dropped
and it hits the surface of water then Y = -5.06m, a= -9.8 and
Vo = 0 at the top. Then using Y = (vf^2 - Vo^2) / 2a i got
Vf = 9.95, then i used that to find out the time and i got
t = 1.01s
Now the total time given is 4.9 so now the time it took for
the ball from the surface of the water to the bottom would be
4.9 -1.01 = 3.89s Now i solved for Y (depth of the lake)
t= 3.89, Vo = 9.95 and Y = ? by using Y = 1/2(Vf+Vo)t and it

2. Jan 25, 2008

### physixguru

For the first problem how can ya take velocity final=0 for the 25 cm jump?

THE APPROACH:

initial velocity=0

dispacement=25 cm = .25m

accleration=g=9.8m/s^2

now use=
s=ut + 1/2at^2 and calculate t.

this gives ya time for bottom 25 cm of jump.

now calculate time taken to reach 55 cm of jump using the above approach.
then calculate v at displacement=55cm[1]>>this becomes u for top 25 cm of jump.

then take final velocity=0
>intial velocity ya got from [1].
>displacement=20cm
>accln=g
calculate t.

FIRST DO THIS PART AND TELL ME IF YA ARE COMFORTABLE..THEN WE SHALL PROCEED WITH NEXT PART>>

3. Jan 26, 2008

### Cathartics

Ok I got t as .22587 for the very first part you mentioned now i am quite confused the way it's laid out. are we working on the problem from the bottom? if so then how can i calculate the time taken to reach 55 cm? I am confused!!

4. Jan 26, 2008

### Cathartics

Someone help me on this one too please!

5. Jan 26, 2008

### chocokat

I'll help, but I'm going to back up a little. The answer you have, t=.22587s is partially correct. But let's look at the problem for a sec.

When the basketball player leaves the floor, he has the fastest velocity, and he slows down (due to gravity) until he reaches the top of the jump, which we know at that point his velocity is zero, and he's 80cm off the ground. Then, as he falls he is speeding up until he hits the floor again.

So, the answer you have, the t=.22587s is actually the time he spends in the air for the last 25cm as he goes up, then another .22587s as he begins to come back down. That makes the time spent in the top 25cm 2 * .22587s.

Now, we can look at the bottom 25cm. It's a little more involved. The way I did the problem was to just look at the descending side. Once again, at the top his velocity is zero (v1 = 0). The total descent is 80cm, so the bottom 25 is from 55-80cm. You can do this a couple of ways. You can calculate the velocity at 55cm, then use that at v1 and calculate the time for the next 25cm. Or, you can calculate the time it takes to get to 55cm (starting at the top) then the time it takes to get to the 80cm (again from the top) and subtract the two times. I think there are other ways to do it. Just remember, the time you calculate is for one way, so you have to double it to get the total time.

You should draw yourself a little picture, a line going up, and a line coming down, then label what you know (which is always more than what is written in the problem, such as gravity, the velocity at the top) and then go from there.

I hope this helps.

6. Jan 26, 2008

### Cathartics

I have already made 4 attempts to this problem and there is one more attempt left... I really dont wanna mess this up if you can please give me the actual answer of (a) in millisecond i am tired of solving this trivial question online.... makes my life miserable....Only this time i will ask for the exact answer i like to get the answer on my own but this time it might cost me so help..

7. Jan 27, 2008

### Cathartics

Thank's chocokat,

I am too scared to attempt problem (a) so i have 3 more attempt for (b) so i tried doing it. Here is what i did. I calculated time at .55m and .80m from the top taking
Vo=o and = -9.8 and i got t1=.3350 and t2 = .4040 thus t2-t1 = .069 and that x2 would be .1380 and it tell me i'm wrong its so frustrating... grrr!! but anyway i'm still trying if you solved the problem can you give me the exact numbers so that i can try match it with mine. Waiting for a quick response and yeah it's due today :( and yeah if you have answer for (a) please give me thanks

Last edited: Jan 27, 2008
8. Jan 27, 2008

### Hootenanny

Staff Emeritus
I'm not quite sure what your doing, but chocokat has given you the answer to (a). For question (b) your asked to calculate the total time spend in the bottom 25cm, this is the sum of the time travelling from 0m to 0.25m and then the time taken to come back down through 0.25m to 0m. However luckily for you, the problem is trivial since the two time intervals are equal and therefore you only have to calculate one of them. So, to start can you calculate the basketball player's intital velocity?

9. Jan 27, 2008

### Cathartics

He gave me the answer for (a) but it says that it's wrong when i plug the number in online.. I dont know why for (b) i get t = .582 and twice the time that is going up and going down would be 1.164 but i am too scared to put it there and i only have one more attempt at it... to get t first i found out Vf from the top where Vo=0 for .55m while coming down and then using that Vf i found time for travel for .25m so is the answer correct or not? Please respond.

10. Jan 27, 2008

### Cathartics

11. Jan 27, 2008

### chocokat

I just did the calculations and I'll help you out. But I'm going to walk you through it to, to hopefully help you understand what I did and why I did it.

So, 0 (zero) is the top and .80m is the bottom. And we know that at the top of the jump (0), the velocity of the jumper is 0 (zero). Acceleration is 9.8 m/s/s in the downward direction, which is also positive for the calculations. The top .25 has been calculated correctly, from 0 to .25m t = .22587s, so 2t = .45175s.

To calculate the bottom .25m, I calculated the time it took to get to .55m, then the time it took to get to .80m (from zero both times), then subtracted the two times.

As before, initial velocity is zero, so the equation ends up as:
$$.55 = \frac{1}{2} 9.8 t^2$$

When I solve for this, the answer is t = .3355s
Then I calculated the time it took to get to .80m:
$$.80 = \frac{1}{2} 9.8 t^2$$
and solving for this gets me .40406s.

To get the time it takes to go from .55m to .80m (the bottom .25m), it's
.40406s - .3355s = .06856s Multiply this by 2, and you get .137s which should be the total time spent in the bottom .25m.

If you think about it, as the person jumps up, he/ has a faster velocity at the bottom and slows down as he approaches the top of the jump. The same happens as he comes down. He is slowest at the top, then speeds up due to the acceleration of gravity. The point of this is that the time spent in the top .25m will be greater than the bottom 25m. It's one of those things to think about as you're getting ready to do the problem, so that you'll have an idea as to whether your calculations are correct.

Please re-do my calculations to be sure they're correct. It's early, and I did them quickly.

12. Jan 27, 2008

### Cathartics

Oh thanks a lot!! thank you thank you thank you so much.. but its asking me the answer in milliseconds! do i divide it by 1000 to get it? also i tried putting .451 for (a) first problem but it says i'm wrong...

13. Jan 27, 2008

### chocokat

To go from seconds to milliseconds, I believe you should multiply by 1000, like this:

$$.45175s * \frac{1000ms}{1s} = 451.75ms$$

I don't know why this answer is wrong, I've done it twice and that's the answer I get.

14. Jan 27, 2008

### Cathartics

ahhh its was my mistake, I was dividing instead of multiplying... Thank you so much chocokat MUAHHHHHHHHHH!!!!! ( i love you)

Last edited: Jan 27, 2008