Calculating Free Fall: Finding Time and Height

[jperk980]

1) A lead ball is dropped into a lake from a diving board 5.12 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 4.72 s after it is dropped. (Assume the positive direction is upward.) How deep is the lake?
For this question I dentified the velocity at which the ball hits the water and that is 10.02 m/s. To get that i used the formula v^2=Vi^2+2ax. I used vi=0, 9.8=a x=5.12. I do not know what to do know because i keep getting the wrong answer and its not off by like like 1 or 2 its off by 10.

2) An object falls from height h from rest and travels 0.46h in the last 1.00 s.
Find the time of its fall.
Find the height of its fall.
For this one I believe i should use the formula x=vt-.5at^2. I know believe time is t-1 and the hieght is h-.46. Could you please help me understand what i should do next becuase i am clueless how to put this information into a formula.
Thank you for the help

 

[Doc Al]

1) A lead ball is dropped into a lake from a diving board 5.12 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 4.72 s after it is dropped. (Assume the positive direction is upward.) How deep is the lake?
For this question I dentified the velocity at which the ball hits the water and that is 10.02 m/s. To get that i used the formula v^2=Vi^2+2ax. I used vi=0, 9.8=a x=5.12.

So far, so good.

I do not know what to do know because i keep getting the wrong answer and its not off by like like 1 or 2 its off by 10.

Start by figuring out how long it takes for the ball to hit the water. Then figure out how much time it spends in the water.

 

[Doc Al]

2) An object falls from height h from rest and travels 0.46h in the last 1.00 s.
Find the time of its fall.
Find the height of its fall.
For this one I believe i should use the formula x=vt-.5at^2. I know believe time is t-1 and the hieght is h-.46.

That will work, using down as positive and thus D = 0.5gt^2. But you need to apply it twice: once for the distance h-.46h and once for the distance h.

Correction: The first position given is .46h, not .46; so the distance fallen is h-.46h.

 

[jperk980]

i don't understand what your saying by using twice how would i put that into a formula and how would i solve for t and h because there would be 2 varibles in that equation you gave me

 

[jperk980]

btw thanks for your help with the first problem it made it a lot easier for me to solve

 

[Doc Al]

i don't understand what your saying by using twice how would i put that into a formula and how would i solve for t and h because there would be 2 varibles in that equation you gave me

Since you have two variables you need two equations. Write one equation for the distance h-.46h & time t-1; write another equation for the distance h and time t. You'll need to solve them together to find h & t.

(See my correction to post #3.)

 

[jperk980]

okay i do that and i come up with the equations h=.5*9.8*t^2 and h-.46h=.5*9.8*(t-1)^2 so now what do i do with that is there a way to set them equal and solve for t because i can't figure out how to do that

 

[Doc Al]

One way to solve these simultaneous equations is to substitute one into the other. Replace "h" in the second equation with .5*9.8*t^2 from the first equation. Then you can solve the quadratic equation for t. (And then plug into get h.)

 

[jperk980]

Thanks Doc Al for all your help. I finally got the answer. Thanks so much.