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AravindG
 4 years ago
two bodies fall freely from different heights and arrive at earth in same time.the time of fall of first body is 2 s and other is 1 s .At what height was first body when second began to fall??
AravindG
 4 years ago
two bodies fall freely from different heights and arrive at earth in same time.the time of fall of first body is 2 s and other is 1 s .At what height was first body when second began to fall??

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AravindG
 4 years ago
Best ResponseYou've already chosen the best response.02.A particle slides from rest from topmost point of a vertical circle of radius r along a smooth chord .Show that the time of descent is same irrespective of chord chosen and find it in terms of g and r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\mathsf{h = u_y\cdot t + \frac{1}{2} \cdot (g) \cdot t^2}\] I am assuming the bodies had Zero initial velocity. \[\mathsf{ h = \frac{1}{2} \cdot g \cdot t^2 }\]I think you can take it from here.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0hey but bodies cover different heights

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For different time. Just put in the value of time for the bodies separately.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0'bodies' sounds a like harsh word for physics. Okay, so we have two Objects A and B. Height or the distance traversed by the Objects is given by the following equation. \[\mathsf{h_{i} = \frac{1}{2}\cdot g \cdot t_{i}^2}\] For the object A, let's assume time taken to land is 2s and for object B 1s. \[\mathsf{h_{A} = \frac{1}{2}\cdot g \cdot t_{A}^2 = \frac{g}{2} \cdot 2^2}\] \[\mathsf{h_{B} = \frac{1}{2}\cdot g \cdot t_{B}^2} = \frac{g}{2} \cdot 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Note. The \(\mathsf{h_{i} = \frac{1}{2}\cdot g \cdot t_{i}^2}\) equation is only valid if the initial velocity of the object 'i' is Zero. \[\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey aravind, do you have a figure (diagram) for the second question?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0pls answr both the qns

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, I thought 1s and 2s is the time taken by each body to land and we have to determine their initial height accordingly. Thanks!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ishaan. The bodies are released at different times. Otherwise, there is no way they could reach the ground at the same time. (At least under Newton's Laws, we won't get into special relativity.) Let's express the height as a function of time and velocity. \[h_1(t,v(t'))=v(t')⋅t+{1 \over 2}⋅g⋅t^2\] The height of the first object at the time the second object is released can be expressed as \[h_1(2,0)−h_1(1,v(1))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I got a sign wrong here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But what about the second question? I don't understand a thing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand it either. A figure would be helpful.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I think it's asking this: along any such chord as in this diagram, the time for the object to reach the the point where the circle and chord meet is equal. Indians ask the damnest questions in Physics. dw:1327424415886:dw

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Hint: this polar form an the equation of a circle or radius R will be helpful: \[ r = 2A\cos \theta \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh I see, thanks James. I think I can do it now.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0**correction: "of radius A" [not R!]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Repost this as a new question. It's worth everyone seeing this one.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0james k can u help me in permutations its urgent??

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study?F434341477489RJUF3Y=_#/updates/4f1ef5f3e4b04992dd24ead0
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