Please help: object falling past window, time given

[jperk980]

Could anyone give me help on these problems thank you.

1. An object falls from height h from rest and travels 0.46h in the last 1.00 s. Find the height and time of its fall. Explain the physically unacceptable solution of the quadratic equation in t that you obtain.
I think this problem uses the equation y=vi*t-.5gt^2, but I am not sure how to use it. Also i have no idea how to even start thinking about answering the last question.

2.A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.37 s, and the top-to-bottom height of the window is 1.65 m. How high above the window top did the flowerpot go?
I know this equation uses the formula y=vi*t-.5gt^2 but I am not sure how to input the information into the formula.

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?What is the average velocity of the ball? 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 4.72 s. What is the initial velocity of the ball?
I do not know were to start on this question.

Thanks for all of your help

 

[learningphysics]

1)... you have two unknowns... the time $$t_{fall}$$ of the fall... and the height... You can use this equation twice:

y=vi*t-.5gt^2 (with vi = 0, since it is just dropped)

You know that in tfall seconds, it drops a height of h (so t=tfall and y = -h)... you also know something about how far it falls in tfall-1 seconds... You'll have two equations with two unknowns h and tfall...

2) Find the velocity when the flower pot goes past the top of the window... hint the time past the window going up, is the same as the time past the window going down. So on the upward path... you know the height, you know the time, you know the acceleration... you can get the velocity at the top...

and when you know that, you can get the height above the top...

3) First find the velocity with which it hits the water... you know the height is 5.12. you know the acceleration... you know it's dropped, so vi=0. that's all you need to get the velocity with which it hits.

Then get the depth of the lake.

Once you know the depth of the lake, then the last part is straightforward...

 

[jperk980]

Thanks learningphysics for the help. I am still having problems with the second question i asked, but i was able to get question 1 and 3 could you please try to explain this concept to me one more time. thank you for all your help

 

[learningphysics]

Sure... the question says that the flower pot was in view for 0.37s... that means that 0.37/2 seconds in view while it was going up... and 0.37/2 seconds in view while going down...
0.37/2 = 0.185s

So examine the upward path while the flower pot is moving from the bottom of the window to the top... the acceleration is g downwards... can you find the velocity at the top of the window or the bottom (either one will help you) ? Use t=0.185s. Look for an equation that will give you what you need...

The way I learned kinematics is with 5 equations:

v2 = v1 + at
d = [(v1+v2)/2]*t
v2^2 = v1^2 + 2as
d = v1*t + (1/2)at^2
d = v2*t - (1/2)at^2 (this one seems to be much less commonly mentioned for some reason)

Try to memorize all 5 of these 5 equations... it really helps to solve this kinematics equations faster...

 

[jperk980]

Okay i did that but how do i get the height, i got the velocity to be 1.08 m/s. once again thanks for the help

 

[learningphysics]

Okay i did that but how do i get the height, i got the velocity to be 1.08 m/s. once again thanks for the help

Can you show how you got this velocity? I get something different.

Once you get the velocity at the bottom or top of the window... then using the fact that the velocity at the top is 0, you can get the distance between these two points... there's a kinematics equation you can use immediately...

 

[jperk980]

im sorry i subtracted instead of adding i got 9.83. I did this by using the formula x=vi*t+.5*a*t^2. I plugged in 1.65=vi(.185)+.5*-4.9*(.185)^2. i got 1.65=vi(.185)-.9065. then i got 1.82=vi(.185) and then vi=9.83. Is that what you get. When i get that am i suppose to use the formula v2 = v1 + at. I used that and get the answer 2.1 but that is not the answer what answer do you get. or what am i doing wrong.

 

[learningphysics]

Yeah, I get vi=9.825

So now, you need the distance to the top... there's a displacement equation you can use to get the displacement from the bottom of the window, to the maximum height... Use vi=9.825, and vf=0.

then subtract the height of the window, because the question asks for the distance from the top of the window...

 

[jperk980]

Thank you so much for all your help. I am starting to get this much better than I did before I asked you. I really appreciate all your time and effort you use to work with me. Thank You Learningphysics!

 

[learningphysics]

Thank you so much for all your help. I am starting to get this much better than I did before I asked you. I really appreciate all your time and effort you use to work with me. Thank You Learningphysics!

You're welcome! Glad to help!

 

[Feldoh]

Could anyone give me help on these problems thank you.

1. An object falls from height h from rest and travels 0.46h in the last 1.00 s. Find the height and time of its fall. Explain the physically unacceptable solution of the quadratic equation in t that you obtain.
I think this problem uses the equation y=vi*t-.5gt^2, but I am not sure how to use it. Also i have no idea how to even start thinking about answering the last question.

If you're in calculus you could also solve by integration. As to why one of the answers is physically unacceptable consider this: can you have negative time?