Solving for Depth, Velocity, and Initial Velocity in a Free Fall Challenge

[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?
Answere in meter

(b) What is the average velocity of the ball?
Answer in m/s

(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?
Answer in m/s

 

[Hootenanny]

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.

 

[Cathartics]

Ok hold on i will post up my attempt

 

[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= volt +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!

 

[Hootenanny]

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

Correct, but be careful of rounding, I have 9.9587...

now i calculated the time using the above also

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

Correct, but again be careful with rounding (t=1.016...)

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= volt +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!

The clue is in the question;

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.

And use 4.92 not just 4.9 for the total time.

 

[Astronuc]

Y= volt +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!

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.

 

[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!

 

[Astronuc]

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)

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

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

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

 

[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 don't 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 don't think so! i tried using Y = volt +1/2 at^2 but i get imaginary roots. so please help on this! once again thank's in advance

 

[Hootenanny]

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 don't know why...

I agree with your answer, except perhaps that is should be negative since the ball is traveling downwards (velocity is a vector quantity).

(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 don't think so! i tried using Y = volt +1/2 at^2 but i get imaginary roots. so please help on this! once again thank's in advance

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.

 

[Cathartics]

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