Homework Help: Another free fall challange

1. Jan 26, 2008

Cathartics

[SOLVED] *****Another free fall challange!!!*****

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?

2. Jan 26, 2008

Hootenanny

Staff Emeritus
Hi Cathartics,

Firstly, homework questions are to be posted in the Homework section. Secondly, one is expected to show an attempt when posting homework questions.

3. Jan 26, 2008

Cathartics

Ok hold on i will post up my attempt

4. Jan 26, 2008

Cathartics

Ok first i found out the velocity it hits the water surface to do that Y= 5.06 and Vo = 0(initial velocity) and Vf = unknown (to be calculated) and a = 9.8 using the formula
Y = Vf^2 - Vo^2/ 2a i get Vf = 9.95 now i calculated the time using the above also

thats Y = 1/2 (Vo + Vf)t and that is t =1.01s

now i can get the time it took from the surface of the water to the bottom by 4.9s-1.01s and that's 3.89s then i can find Y from the surface of the water to the bottom by using
Y= Vot +1/2 at^2 where Y = unknown and Vo= 9.95 and t = 3.89 and it comes in quadratic which would give me Y = 19.59 and the answer is wrong now please help me!!!!

5. Jan 26, 2008

Hootenanny

Staff Emeritus
Correct, but be careful of rounding, I have 9.9587...
Correct, but again be careful with rounding (t=1.016...)
The clue is in the question;
And use 4.92 not just 4.9 for the total time.

Last edited: Jan 26, 2008
6. Jan 26, 2008

Astronuc

Staff Emeritus
The formula is not correct. The formula assumes a constant acceleration (or deceleration).

The problem statement states "then sinks to the bottom with this same constant velocity", in which case h = v*t.

I see Hoot has responded also.

7. Jan 26, 2008

Cathartics

Thanks hoot and Astronuc for the response.

So astronuc you mean to say if i do h = 9.95 x 3.89 i will get the answer? for (a)

and how to you calculate the average velocity? and how do you do the part (c) i'm lost please help!

8. Jan 26, 2008

Astronuc

Staff Emeritus
for constant velocity, the distance (or height) is just velocity * time. So, yes.

For b) one simple way to determine the average velocity (speed) is to take the total distance and divide by total time to traverse the distance. That defines the 'average' velocity, as opposed to instanteous velocity, which changes under the constant acceleration of free-fall.

in c) What is total distance the ball falls from board to lake bottom? One is given the time. So write the equation for a downward trajectory with an initial velocity.

Please refer to this site for equations of motion.
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

9. Jan 27, 2008

Cathartics

Dear Astronuc,

Thanks for that info i got (a) correct ***yehhhh*** now i'm trying to do (b)
so you said if i take the total distance i.e 5.06 plus the height of the lake that's 38.739400 that would be 43.7994 and now divide this by the total time that's 4.92 i would get 8.90m/s but it says i'm wrong dont know why...

(c) ok for this i know Y= 43.7994 and time t = 4.92 but they are asking to find the initial velocity but do we know the final velocity?i dont think so!! i tried using Y = vot +1/2 at^2 but i get imaginary roots. so please help on this! once again thank's in advance

10. Jan 27, 2008

Hootenanny

Staff Emeritus
I agree with your answer, except perhaps that is should be negative since the ball is travelling downwards (velocity is a vector quantity).
How are you getting imaginary roots? Your solving for v_0, not t. You should also be aware that both Y and a are negative quantities.

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

Cathartics

Thank you Hoot, Thank you thank you thank you so much!! appreciate it...

Last edited: Jan 27, 2008